Revision Notes for CBSE Class 12 Maths Chapter 8 (Application of Integrals) - Free PDF Download
Section–A (1 Mark Questions)
1. Find the area of the region bounded by the curve $y=\sin x$ between the ordinates $x=0, x=\frac{\pi}{2}$ and the $x$-axis.
Ans. We have $y=\sin x, 0 \leq x \leq \frac{\pi}{2}$
From the figure, area of the shaded region
$\begin{aligned}& A=\int_0^{\pi / 2} \sin x d x=[-\cos x]_0^{\pi / 2} \\& =-\cos \frac{\pi}{2}+\cos 0=0+1=1 \text { sq. units }\end{aligned}$
2. Find the area of the region bounded by the curve $y=x+1$ and the lines $x=2$ and $x=3$.
Ans. From the figure, area of the shaded region,
$$\begin{aligned}& A=\int_2^3(x+1) d x=\left[\frac{x^2}{2}+x\right]_2^3 \\& =\left[\frac{9}{2}+3-\frac{4}{2}-2\right] \\& =\frac{7}{2} \text { sq. units }\end{aligned}$$
3. Find the area of the region bounded by the curve $x=2 y+3$ and the line $y=1$ and $y=-1$.
Ans. From the figure, area of the shaded region,
$$\begin{aligned}& A=\int_{-1}^1(2 y+3) d y=\left[y^2+3 y\right]_{-1}^1 \\& =[1+3-1+3]=6 \text { sq. units }\end{aligned}$$
4. Find the area of the region bounded by the curve $x=y^2, y$ - axis and the lines $$y=3 \text { and } y=4 .$$
Ans. $$\begin{aligned}& \text { Required area, } A=\int_3 x d y \\& =\int_3^4 y^2 d y \\& =\left[\frac{y^3}{3}\right]_3^4 \\& =\frac{1}{3}\left[4^3-3^3\right] \\& =\frac{37}{3} \text { sq. units }\end{aligned}$$
5. Using integration, find the area of the region bounded by the line $2 y=5 x+7$, $x$-axis and the lines $x=2$ and $x=8$.
Ans. We have, $2 y=5 x+7$ Or $y=\frac{5 x}{2}+\frac{7}{2}$
The graph is as shown in figure below. From the figure, area of shaded region
$$\begin{aligned}& =\int_2^8 \frac{5 x+7}{2} d x \\& =\frac{1}{2}\left[5\left(\frac{x^2}{2}\right)+7 x\right]_2^8 \\& =\frac{1}{2}[5(32)+7(8)-10-14] \\& =\frac{1}{2}[160+56-24] \\& =96 \text { sq. units. }\end{aligned}$$
Section-B (2 Marks Questions)
6. Find the area of the region bounded by the curve $y=\cos x$ between $x=0$ and $x=\pi$.
Ans. We have $y=\cos x, x=0$ and $x=\pi$
From the figure, area of the shaded region,
$$\begin{aligned}& A=\int_0^{\pi / 2} \cos x d x+\int_{\pi / 2}^\pi|\cos x| d x \\& =2 \int_0^{\frac{\pi}{2}} \cos x d x=2[\sin x]_0^{\pi / 2}=2 \text { sq. units. }\end{aligned}$$
7. Find the area bounded by the parabola $y^2=4 x$ and line $x=3$.
Ans. Required Area $=$ area $\mathrm{AOBMA}=2 \times \operatorname{area}($ AOMA $)$ ($\because$ Parabola is symmetrical about $x$-axis)
$$\begin{aligned}& =2 \int_0^a y d x=2 \int_0^a \sqrt{4 a x} d x \\& =2 \times 2 \sqrt{a} \int_0^a \sqrt{x} d x=4 \sqrt{a} \times\left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^a \\& =\frac{8 \sqrt{a}}{3}\left[a^{3 / 2}\right]=\frac{8}{3} a^2 \quad \text { sq. units. }\end{aligned}$$sq. units. I
8. Find the area of the region bounded by the parabolas $y^2=6 x$ and $x^2=6 y$.
Ans. The intersecting points of the given parabolas are obtained by solving these equations for $x$ and $y$, which are $O(0,0)$ and $(6,6)$.
$$\begin{aligned}& \text { Area } O A B C O=\int_0^6\left(\sqrt{6 x}-\frac{x^2}{6}\right) d x \\& =\left[2 \sqrt{6} \frac{x^{\frac{3}{2}}}{3}-\frac{x^3}{18}\right]_0^6 \\& =2 \sqrt{6} \frac{(6)^{\frac{3}{2}}}{3}-\frac{(6)^3}{18}=12 \text { sq. units. }\end{aligned}$$
9. The area of the region bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is
Ans. We have, $\frac{x^2}{25}+\frac{y^2}{16}=1$ which is ellipse with its axis as coordinate axis.
$$\begin{aligned}& \frac{y^2}{16}=1-\frac{x^2}{25} \Rightarrow y^2=16\left(1-\frac{x^2}{25}\right) \\& \Rightarrow y=\frac{4}{5} \sqrt{5^2-x^2}\end{aligned}$$
From the figure, area of the shaded region.
$$\begin{aligned}& A=4 \int_0^5 \frac{4}{5} \sqrt{5^2-x^2} d x \\& =\frac{16}{5}\left[\frac{x}{2} \sqrt{5^2-x^2}+\frac{5^2}{2} \sin^{-1}\left(\frac{x}{5}\right)\right]_0^5\end{aligned}$$
$$\begin{aligned}& =\frac{16}{5}\left[0+\frac{5^2}{2} \sin ^{-1}(1)-0-0\right] \\& =\frac{16}{5} \cdot \frac{25}{2} \cdot \frac{\pi}{2}=20 \pi \text { sq. units. }\end{aligned}$$
10. Find the area of the parabola $y^2=4 a x$ bounded by its latus rectum.
Ans. The vertex of the parabola $y^2=4 a x$ is at origin $(0,0)$. The equation of the latus rectum LSL' is $x=a$. Also, parabola is symmetrical about the x-axis.
The required area of the region $O L L^{\prime} \mathrm{O}$ $=2 \times($ Area of the region $\mathrm{O} L S O)$.
$=2 \int_0^a y d x=2 \int_0^a \sqrt{4 a x} d x$
$=2 \times 2 \sqrt{a} \int_0^a \sqrt{x} d x=4 \sqrt{a} \times\left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^a$
$=\frac{8 \sqrt{a}}{3}\left[a^{3 / 2}\right]=\frac{8}{3} a^2 \quad \text { sq. units. }$
11. Find the area enclosed by parabola $y=4 x^2$ and lines, $y=1$ and $y=4$.
Ans. Parabola $y=4 x^2$
So $x^2=\frac{1}{4} y$ and lines $y=1$ and $y=4$ will be traced as followed.
So, required area $=$ area $P Q R S P$
$$\begin{aligned}& =2 \times \operatorname{Area}(\text { RQLMR }) \\& =2 \int_1^4 x d y=2 \int_1^4 \frac{1}{2} \sqrt{y} d y=\int_1^4 \sqrt{y} d y \\& =\frac{2}{3}\left[(y)^{3 / 2}\right]_1^4=\frac{2}{3}\left[4^{3 / 2}-1^{3 / 2}\right] \\& =\frac{2}{3}[8-1]=\frac{14}{3} \text { sq. units. }\end{aligned}$$
12. Find the area enclosed by the curve $y^{y=-x^2}$ and the straight line $x+y+2=0$.
Ans. We have, $y=-x^2$ and $x+y+2=0$
Solving we get, $$\begin{aligned}& x^2=x+2 \\& \Rightarrow x^2-x-2=0 \\& \Rightarrow(x-2)(x+1)=0 \\& \Rightarrow x=2,-1\end{aligned}$$
The graph of above function is downward parabola. From the figure, area of shaded region,
$A=\int_{-1}^2\left(-x^2-(-x-2)\right) d x$
$=\int_{-1}^2\left(x+2-x^2\right) d x$
$=\left[\frac{x^2}{2}+2 x-\frac{x^3}{3}\right]_{-1}^2$
$=\left[\frac{2^2}{2}+2(2)-\frac{2^3}{3}\right]-\left [ \frac{(-1)^{2}}{2}+(-1)-\frac{(-1)^{3}}{3} \right ]$
$=\left[6-\frac{8}{3}\right]-\left[\frac{1}{2}-2+\frac{1}{3}\right]$
$=\frac{9}{2} \text { sq. units }$
13. Find the area bounded by the curve $y=\sqrt{x}, \quad x=2 y+3$ in the first quadrant and $x$-axis.
Ans. We have $y=\sqrt{x}$ and $x=2 y+3$
Solving we get
$$ \begin{aligned} & y=\sqrt{2 y+3}, y \geq 0 \\ & \Rightarrow y^2=2 y+3, y \geq 0 \\ & \Rightarrow y^2-2 y-3=0, y \geq 0 \\ & \Rightarrow(y-3)(y+1)=0, y \geq 0 \\ & \Rightarrow y=3 \end{aligned} $$
The graph of function $y=\sqrt{x}$ is part of parabola $y^2=x$ lying above $x$-axis. The graph is as shown in the following figure.
From the figure, area of shaded region,
$$ \begin{aligned} & A=\int_0^3\left(2 y+3-y^2\right) d y \\ & =\left[y^2+3 y-\frac{y^3}{3}\right]_0^3 \\ & =[9+9-9-0] \\ & =9 \text { sq. units. } \end{aligned} $$
PDF Summary - Class 12 Maths Application of Integrals Notes (Chapter 8)
Definite Integration
1. Definition:
If \[\text{ }\!\!~\!\!\text{ F}\left( \text{x} \right)\] is an antiderivative of \[\text{f}\left( \text{x} \right)\text{,}\] then \[\text{ }\!\!~\!\!\text{ F}\left( \text{b} \right)\text{--F}\left( \text{a} \right)\] is known as the definite integral of \[\text{f}\left( \text{x} \right)\] from $\text{a}$ to $\text{b}$, such that the variable \[\text{x,}\] takes any two independent values say \[\text{a}\] and \[\text{b}\].
This is also denoted as $\int_{\text{a}}^{\text{b}}{\text{f(x)dx}}$.
Thus $\int_{\text{a}}^{\text{b}}{\text{f(x)dx}}=\text{F}\left( \text{b} \right)-\text{F}\left( \text{a} \right)$, The numbers $\text{a}$ and $\text{b}$ are called the limits of integration; $\text{a}$ is the lower limit and $\text{b}$ is the upper limit. Usually $\text{F}\left( \text{b} \right)-\text{F}\left( \text{a} \right)$ is abbreviated by writing $\text{F}\left( \text{x} \right)\left| _{\text{a}}^{\text{b}} \right.$.
2. Properties of Definite Integrals:
l. $\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx=}-\int\limits_{\text{b}}^{\text{a}}{\text{f(x)}}}$
ll. $\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx=}\int\limits_{\text{b}}^{\text{a}}{\text{f(y)}}}\text{dy}$
lll. $\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx=}\int\limits_{\text{a}}^{\text{c}}{\text{f(x)dx+}\int\limits_{\text{c}}^{\text{b}}{\text{f(x)dx}}}}$, where $\text{c}$ may or may not lie between $\text{a}$ and $\text{b}$.
lV. $\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx=}\int\limits_{\text{0}}^{\text{a}}{\text{f(a}-\text{x)dx}}}$
V. $\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx=}\int\limits_{\text{a}}^{\text{b}}{\text{f(a+b}-\text{x)dx}}}$
Note:
$\int\limits_{\text{0}}^{\text{a}}{\dfrac{\text{f(x)}}{\text{f(x)+f(a}-\text{x)}}}\text{dx=}\dfrac{\text{a}}{\text{2}}$
$\int\limits_{\text{a}}^{\text{b}}{\dfrac{\text{f(x)}}{\text{f(x)+f(a+b}-\text{x)}}}\text{dx=}\dfrac{\text{b}-\text{a}}{\text{2}}$
Vl. $\int\limits_{\text{0}}^{\text{2a}}{\text{f(x)dx=}\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx+}\int\limits_{\text{0}}^{\text{a}}{\text{f(2a}-\text{x)dx}}}}$
$\text{=}\left\{ \begin{matrix} \text{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{if}\,\text{f(2a}-\text{x)}=-\text{f(x)} \\ \text{2}\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx}\,\,\,\,\,\,\,\,\,\,\,\,\text{if}\,\text{f(2a}-\text{x)}=\text{f(x)}} \\ \end{matrix} \right\}$
Vll. \[\int\limits_{\text{-a}}^{\text{a}}{\text{f(x)dx}=\left\{ \begin{matrix} \text{2}\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx}\,\,\,\,\,\,\,\,\,\,\text{if}\,\text{f(}-\text{x)}=\text{f(x)}\,\text{i}\text{.e}\text{.}\,\text{f(x)}\,\text{is}\,\text{even}} \\ \text{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{if}\,\text{f(}-\text{x)}=-\text{f(x)}\,\text{i}\text{.e}\text{.}\,\text{f(x)}\,\text{is}\,\text{odd} \\ \end{matrix} \right\}}\]
Vlll. If $\text{f(x)}$ is a periodic function of period $\text{ }\!\!'\!\!\text{ a }\!\!'\!\!\text{ }$, i.e. $\text{f(a}+\text{x)}=\text{f(x)}$, then
$\int\limits_{\text{0}}^{\text{na}}{\text{f(x)dx}=\text{n}\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx}}}$
$\int\limits_{\text{0}}^{\text{na}}{\text{f(x)dx}=\left( \text{n}-1 \right)\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx}}}$
$\int\limits_{\text{na}}^{\text{b+na}}{\text{f(x)dx}=\int\limits_{\text{0}}^{\text{b}}{\text{f(x)dx}}}$, where $\text{b}\in \text{R}$
$\int\limits_{\text{b}}^{\text{b+a}}{\text{f(x)dx}}$ independent of $\text{b}$.
$\int\limits_{\text{b}}^{\text{b+na}}{\text{f(x)dx}=\text{n}\int\limits_{\text{0}}^{\text{a}}{\text{f(x)dx}}}$, where $\text{n}\in \text{1}$
lX. If $\text{f(x)}\ge \text{0}$ on the interval $\left[ \text{a,b} \right]$, then $\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}}\ge 0$.
X. If $\text{f(x)}\le \text{g(x)}$ on the interval $\left[ \text{a,b} \right]$, then $\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}}\le \int\limits_{\text{a}}^{\text{b}}{\text{g(x)dx}}$
X. $\left| \int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}} \right|\le \int\limits_{\text{a}}^{\text{b}}{\left| \text{f(x)} \right|\text{dx}}$
Xl. If $\text{f(x)}$ is continuous on $\left[ \text{a,b} \right]$, $\text{m}$ is the least and $\text{M}$ is the greatest value of $\text{f(x)}$ on $\left[ \text{a,b} \right]$, then
$\text{m}\left( \text{b}-\text{a} \right)\le \int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}}\le \text{M(b}-\text{a)}$
Xll. For any two functions $\text{f(x)}$ and $\text{g(x)}$, integral on the interval $\left[ \text{a,b} \right]$, the Schwarz- Bunyakovsky inequality holds
$\left| \int\limits_{\text{a}}^{\text{b}}{\text{f(x)}\text{.g(x)dx}} \right|\le \sqrt{\int\limits_{\text{a}}^{\text{b}}{{{\text{f}}^{\text{2}}}\text{(x)dx}\text{.}\int\limits_{\text{a}}^{\text{b}}{{{\text{g}}^{\text{2}}}\text{(x)dx}}}}$
Xlll. If a function $\text{f(x)}$ is continuous on the interval $\left[ \text{a,b} \right]$, then there exists a point $\text{c}\in \left( \text{a,b} \right)$ such that
\[\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}=\text{f(c)(b}-\text{a)}}\], where $\text{a}<\text{c}<\text{b}$.
3. Differentiation Under Integral Sign:
Newton Leibnitz’s Theorem:
Given that $\text{x}$ has two differentiable functions, $\text{g(x)}$ and $\text{h(x)}$, where $\text{x}\in \left[ \text{a,b} \right]$ and $\text{f}$ is continuous in that interval, then
$\dfrac{\text{d}}{\text{dx}}\left[ \int\limits_{\text{g(x)}}^{\text{h(x)}}{\text{f(t)dt}} \right]=\dfrac{\text{d}}{\text{dx}}\left[ \text{h(x)} \right]\text{.f}\left[ \text{h(x)} \right]-\dfrac{\text{d}}{\text{dx}}\left[ \text{g(x)} \right]\text{.f}\left[ \text{g(x)} \right]$
4. Definite Integral as a Limit of Sum:
Let $\text{f(x)}$ be a continuous real valued function defined on the closed interval $\left[ \text{a,b} \right]$ which is divided into $\text{n}$ parts as shown in figure.
The point of division on $\text{x-}$axis are
$\text{a,a}+\text{h,a}+\text{2h}.....\text{a}+\left( \text{n}-\text{1} \right)\text{h,a}+\text{nh,}$ where $\dfrac{\text{b}-\text{a}}{\text{n}}=\text{h}$.
Let ${{\text{S}}_{\text{n}}}$ denotes the area of these $\text{n}$ rectangles.
Then, ${{\text{S}}_{\text{n}}}=\text{hf(a)}+\text{hf}\left( \text{a}+\text{h} \right)+\text{hf}\left( \text{a}+\text{2h} \right)+....+\text{hf}\left( \text{a}+\left( \text{n}-\text{1} \right)\text{h} \right)$
Clearly, ${{\text{S}}_{\text{n}}}$ is area very close to the area of the region bounded by
Curve $\text{y}=\text{f(x),}$ $\text{x-}$axis and the ordinates $\text{x}=\text{a}$, $\text{x}=\text{b}$.
Hence \[\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}=\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,}{{\text{S}}_{\text{n}}}\]
\[\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}=\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,}\sum\limits_{\text{r}=\text{0}}^{\text{n-1}}{\text{hf(a+rh)}}\]
\[=\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\sum\limits_{\text{r}=\text{0}}^{\text{n-1}}{\left( \dfrac{\text{b}-\text{a}}{\text{n}} \right)}\text{f}\left( \text{a}+\dfrac{\left( \text{b}-\text{a} \right)\text{r}}{\text{n}} \right)\]
Note:
(a) We can also write
${{\text{S}}_{\text{n}}}=\text{hf}\left( \text{a}+\text{h} \right)+\text{hf}\left( \text{a}+\text{2h} \right)+.....+\text{hf}\left( \text{a}+\text{nh} \right)$
\[\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}}=\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\sum\limits_{\text{r}=1}^{\text{n}}{\left( \dfrac{\text{b}-\text{a}}{\text{n}} \right)}\text{f}\left( \text{a}+\left( \dfrac{\text{b}-\text{a}}{\text{n}} \right)\text{r} \right)\]
(b) If \[\text{a}=0,\,\text{b}=1,\,\int\limits_{0}^{1}{\text{f(x)dx}}=\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\sum\limits_{\text{r}=0}^{\text{n}-1}{\dfrac{1}{\text{n}}\text{f}\left( \dfrac{\text{r}}{\text{n}} \right)}\]
Steps to Express the Limit of Sum as Definite Integral
Step 1. Replace \[\dfrac{\text{r}}{\text{n}}\] by $\text{x}$, $\dfrac{\text{1}}{\text{n}}$ by $\text{dx}$ and $\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\sum{\text{by}\,}\int{{}}$
Step 2. Evaluate $\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\left( \dfrac{\text{r}}{\text{n}} \right)$ by putting least and greatest values of $\text{r}$ as lower and upper limits respectively.
For example $\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\sum\limits_{\text{r}=1}^{\text{pn}}{\dfrac{1}{\text{n}}\text{f}\left( \dfrac{\text{r}}{\text{n}} \right)}=\int\limits_{0}{\text{f(x)dx}}$
$\left[ \underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\,\left( \dfrac{\text{r}}{\text{n}} \right)\left| _{\text{r}=1}=0,\,\underset{\text{n}\to \infty }{\mathop{\text{Lt}}}\, \right.\left( \dfrac{\text{r}}{\text{n}} \right)\left| _{\text{r}=\text{np}}=\text{p} \right. \right]$
5. Reduction Formulae in Definite Integrals
l. If ${{\text{I}}_{\text{n}}}\text{=}\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{n}}}\text{xdx}}$, then show that ${{\text{I}}_{\text{n}}}\text{=}\left( \dfrac{\text{n}-\text{1}}{\text{n}} \right){{\text{I}}_{\text{n}-\text{2}}}$
Proof: ${{\text{I}}_{\text{n}}}\text{=}\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{n}}}\text{xdx}}$
${{\text{I}}_{\text{n}}}\text{=}\left[ -\text{si}{{\text{n}}^{\text{n}-\text{1}}}\text{xcosx} \right]_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}\text{+}\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\left( \text{n}-\text{1} \right)\text{si}{{\text{n}}^{\text{n}-\text{2}}}\text{x}\text{.co}{{\text{s}}^{\text{2}}}\text{xdx}}$
$\text{=}\left( \text{n}-\text{1} \right)\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{n}-\text{2}}}\text{x}\text{.}\left( \text{1}-\text{si}{{\text{n}}^{\text{2}}}\text{x} \right)\text{dx}}$
$\text{=}\left( \text{n}-\text{1} \right)\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{n}-\text{2}}}\text{xdx}-\left( \text{n}-1 \right)\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{n}}}\text{x}}\text{dx}}$
${{\text{I}}_{\text{n}}}+\left( \text{n}-\text{1} \right){{\text{I}}_{\text{n}}}=\left( \text{n}-\text{1} \right){{\text{I}}_{\text{n}-\text{2}}}$
${{\text{I}}_{\text{n}}}=\left( \dfrac{\text{n}-\text{1}}{\text{n}} \right){{\text{I}}_{\text{n}-\text{2}}}$
Note:
(a) $\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{n}}}\text{xdx}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{co}{{\text{s}}^{\text{n}}}\text{xdx}}}$
(b) ${{\text{I}}_{\text{n}}}=\left( \dfrac{\text{n}-\text{1}}{\text{n}} \right)\left( \dfrac{\text{n}-\text{3}}{\text{n}-\text{2}} \right)\left( \dfrac{\text{n}-\text{5}}{\text{n}-\text{4}} \right)....{{\text{I}}_{\text{0}}}\,\text{or}\,{{\text{I}}_{\text{1}}}$ according as \[\text{n}\] is even or odd, ${{\text{I}}_{\text{0}}}=\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\,{{\text{I}}_{\text{1}}}=\text{1}$
Hence ${{\text{I}}_{\text{n}}}=\left\{ \begin{matrix} \left( \dfrac{\text{n}-\text{1}}{\text{n}} \right)\left( \dfrac{\text{n}-\text{3}}{\text{n}-\text{2}} \right)\left( \dfrac{\text{n}-\text{5}}{\text{n}-\text{4}} \right).....\left( \dfrac{\text{1}}{\text{2}} \right)\text{.}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,\,\text{if}\,\,\text{n}\,\,\text{is}\,\,\text{even} \\ \left( \dfrac{\text{n}-\text{1}}{\text{n}} \right)\left( \dfrac{\text{n}-\text{3}}{\text{n}-\text{2}} \right)\left( \dfrac{\text{n}-\text{5}}{\text{n}-\text{4}} \right).....\left( \dfrac{\text{2}}{\text{3}} \right)\text{.1}\,\,\text{if}\,\,\text{n}\,\,\text{is }\,\text{odd} \\ \end{matrix} \right\}$
ll. If ${{\text{I}}_{\text{n}}}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{ta}{{\text{n}}^{\text{n}}}\text{xdx}}$, then show that ${{\text{I}}_{\text{n}}}+{{\text{I}}_{\text{n}-\text{2}}}=\dfrac{\text{1}}{\text{n}-\text{1}}$
Proof: ${{\text{I}}_{\text{n}}}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{{{\left( \text{tanx} \right)}^{\text{n}-\text{2}}}\text{.ta}{{\text{n}}^{\text{2}}}\text{xdx}}$
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{{{\left( \text{tanx} \right)}^{\text{n}-\text{2}}}\left( \text{se}{{\text{c}}^{\text{2}}}\text{x}-\text{1} \right)\text{dx}}$
$\text{=}\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{{{\left( \text{tanx} \right)}^{\text{n}-\text{2}}}\text{se}{{\text{c}}^{\text{2}}}\text{xdx}}-\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{{{\left( \text{tanx} \right)}^{\text{n}-\text{2}}}\text{dx}}$
$=\left[ \dfrac{{{\left( \text{tanx} \right)}^{\text{n}-\text{1}}}}{\text{n}-\text{1}} \right]_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}-{{\text{I}}_{\text{n}-\text{2}}}$
${{\text{I}}_{\text{n}}}=\dfrac{\text{1}}{\text{n}-\text{1}}-{{\text{I}}_{\text{n}-\text{2}}}$
${{\text{I}}_{\text{n}}}+{{\text{I}}_{\text{n}-\text{2}}}=\dfrac{\text{1}}{\text{n}-\text{1}}$
lll. If ${{\text{I}}_{\text{m,n}}}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{m}}}\text{x}\text{.co}{{\text{s}}^{\text{n}}}\text{xdx}}$, then show that ${{\text{I}}_{\text{m,n}}}=\dfrac{\text{m}-\text{1}}{\text{m}+\text{n}}{{\text{I}}_{\text{m}-\text{2}}}\text{,n}$
Proof: ${{\text{I}}_{\text{m,n}}}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{m}-1}}\text{x}\left( \text{sinxco}{{\text{s}}^{\text{n}}}\text{x} \right)\text{dx}}$
$=\left[ -\dfrac{\text{si}{{\text{n}}^{\text{m}-\text{1}}}\text{x}\text{.co}{{\text{s}}^{\text{n}+\text{1}}}\text{x}}{\text{n}+\text{1}} \right]_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}+\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\dfrac{\text{co}{{\text{s}}^{\text{n}+\text{1}}}}{\text{n}+\text{1}}\left( \text{m}-\text{1} \right)\text{si}{{\text{n}}^{\text{m}-\text{2}}}\text{xcosxdx}}$
$=\left( \dfrac{\text{m}-\text{1}}{\text{n}+\text{1}} \right)\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{m}-\text{2}}}\text{x}\text{.co}{{\text{s}}^{\text{n}}}\text{x}\text{.co}{{\text{s}}^{\text{2}}}\text{xdx}}$
$=\left( \dfrac{\text{m}-\text{1}}{\text{n}+\text{1}} \right)\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\left( \text{si}{{\text{n}}^{\text{m}-\text{2}}}\text{x}\text{.co}{{\text{s}}^{\text{n}}}\text{x}-\text{si}{{\text{n}}^{\text{m}}}\text{x}\text{.co}{{\text{s}}^{\text{n}}}\text{x} \right)\text{dx}}$
$=\left( \dfrac{\text{m}-\text{1}}{\text{n}+\text{1}} \right){{\text{I}}_{\text{m}-\text{2,n}}}-\left( \dfrac{\text{m}-\text{1}}{\text{n}+\text{1}} \right){{\text{I}}_{\text{m,n}}}$
$\Rightarrow \left( \text{1}+\dfrac{\text{m}-\text{1}}{\text{n}+\text{1}} \right){{\text{I}}_{\text{m,n}}}=\left( \dfrac{\text{m}-\text{1}}{\text{n}+\text{1}} \right){{\text{I}}_{\text{m}-\text{2,n}}}$
${{\text{I}}_{\text{m,n}}}\text{=}\left( \dfrac{\text{m}-\text{1}}{\text{m}+\text{n}} \right){{\text{I}}_{\text{m}-\text{2,n}}}$
Note:
(a) ${{\text{I}}_{\text{m,n}}}=\left( \dfrac{\text{m}-\text{1}}{\text{m}+\text{n}} \right)\left( \dfrac{\text{m}-\text{3}}{\text{m}+\text{n}-\text{2}} \right)\left( \dfrac{\text{m}-\text{5}}{\text{m}+\text{n}-\text{4}} \right).....{{\text{I}}_{\text{0,n}}}\,\text{or}\,{{\text{I}}_{\text{1,n}}}$ according as $\text{m}$ is even or odd.
${{\text{I}}_{\text{0,n}}}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{co}{{\text{s}}^{\text{n}}}\text{xdx}\,\text{and}\,{{\text{I}}_{\text{1,n}}}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{sinx}\text{.co}{{\text{s}}^{\text{n}}}\text{xdx}=\dfrac{\text{1}}{\text{n+1}}}}$
(b) Walli’s Formula
\[{{\text{I}}_{\text{m,n}}}=\left\{ \begin{matrix} \dfrac{\left( \text{m}-\text{1} \right)\left( \text{m}-\text{3} \right)\left( \text{m}-\text{5} \right)....\left( \text{n}-\text{1} \right)\left( \text{n}-\text{3} \right)\left( \text{n}-\text{5} \right)....}{\left( \text{m}+\text{n} \right)\left( \text{m}+\text{n}-\text{2} \right)\left( \text{m}+\text{n}-\text{4} \right).....}\,\,\,\text{when}\,\,\text{both}\,\,\text{m,n}\,\,\text{are}\,\,\text{even} \\ \dfrac{\left( \text{m}-\text{1} \right)\left( \text{m}-\text{3} \right)\left( \text{m}-\text{5} \right)....\left( \text{n}-\text{1} \right)\left( \text{n}-\text{3} \right)\left( \text{n}-\text{5} \right).....}{\left( \text{m}+\text{n} \right)\left( \text{m}+\text{n}-\text{2} \right)\left( \text{m}+\text{n}-\text{4} \right)....}\,\,\text{otherwise} \\ \end{matrix} \right\}\]
Area Under the Curves
(a) Area of Plane Regions:
l. The area bounded by the curve $\text{y}=\text{f(x)}$, $\text{x-}$axis and the ordinates $\text{x}=\text{a}$ and \[\text{x}=\text{b}\] (where $\text{ba}$) is given by
$\text{A}=\int\limits_{\text{a}}^{\text{b}}{\left| \text{y} \right|\text{dx}=\int\limits_{\text{a}}^{\text{b}}{\left| \text{f(x)} \right|\text{dx}}}$
(i) If $\text{f(x)}>0\forall \text{x}\in \left[ \text{a,b} \right]$
Then $\text{A}=\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}}$
(ii) If $\text{f(x)}>0\forall x\in \left[ \text{a,c} \right)\,\text{ }\!\!\And\!\!\text{ }\,<0\forall \text{x}\in \left( \text{c,b} \right]$
Then $\text{A}=\left| \int\limits_{\text{a}}^{\text{c}}{\text{ydx}} \right|+\left| \int\limits_{\text{c}}^{\text{b}}{\text{ydx}} \right|=\int\limits_{\text{a}}^{\text{c}}{\text{f(x)dx}-\int\limits_{\text{c}}^{\text{b}}{\text{f(x)dx}}}$ where $\text{c}$ is a point in between $\text{a}$ and $\text{b}$.
ll. The area bounded by the curve $\text{x}=\text{g(y)}$, $\text{y-}$axis and the abscissae $\text{y}=\text{c}$ and $\text{y}=\text{d}$ (where $\text{dc}$) is given by
$\text{A}=\int\limits_{\text{c}}^{\text{d}}{\left| \text{x} \right|\text{dy}=\int\limits_{\text{c}}^{\text{d}}{\left| \text{g(y)} \right|\text{dy}}}$
III. If we have two curve $\text{y}=\text{f(x)}$ and $\text{y}=\text{g(x)}$, such that $\text{y}=\text{f(x)}$ lies above the curve $\text{y}=\text{g(x)}$ then the area bounded between them and the ordinates $\text{x}=\text{a}$ and $\text{x}=\text{b}$ $\left( \text{b}>\text{a} \right)$, is given by
$\text{A}=\int\limits_{\text{a}}^{\text{b}}{\text{f(x)dx}-\int\limits_{\text{a}}^{\text{b}}{\text{g(x)dx}}}$
i.e. $\text{upper}\,\text{curve}\,\text{area}-\text{lower}\,\text{curve}\,\text{area}$.
lV. The area bounded by the curves $\text{y}=\text{f(x)}$ and $\text{y}=\text{g(x)}$ between the ordinates $\text{x}=\text{a}$ and $\text{x}=\text{b}$ is given by
$\text{A}=\int\limits_{\text{a}}^{\text{c}}{\text{f(x)dx}+\int\limits_{\text{c}}^{\text{b}}{\text{g(x)dx}}}$, where $\text{x}=\text{c}$ is the point of intersection of the two curves.
V. Curve Tracing
It is necessary to have a rough sketch of the desired piece in order to locate the area enclosed by many curves. The following steps are very useful in tracing a cartesian curve $\text{f(x,y)}=\text{0}$.
Step 1: Symmetry
(i) If all the powers of $\text{y}$ in the equation of the given curve are even then the curve is symmetrical about $\text{x-}$axis.
(ii) If all the powers of $\text{x}$ in the equation of the given curve are even then the curve is symmetrical about $\text{y-}$axis.
(iii) If the equation of the given curve remains unchanged on interchanging $\text{x}$ and $\text{y}$, then the curve is symmetrical about the line\[\text{y}=\text{x}\].
(iv) If the equation of the given curve remains unchanged when $\text{x}$ and $\text{y}$ are replaced by $-\text{x}$ and $-\text{y}$ respectively, then the curve is symmetrical in opposite quadrants.
Step 2: Origin
If the algebraic curve's equation contains no constant term, the curve passes through the origin.
The tangents at the origin are then calculated by equating the lowest degree terms in the equation of the specified algebraic curve to zero.
For example, the curve ${{\text{y}}^{\text{3}}}={{\text{x}}^{\text{3}}}+\text{axy}$ passes through the origin and the tangents at the origin are given by $\text{axy}=0$ i.e. $\text{x}=\text{0}$ and $\text{y}=0$.
Step 3: Intersection with the Co-ordinates Axes
(i) Find out the corresponding values of $\text{x}$ by putting \[\text{y}=\text{0}\] in the equation of the given curve, to estimate the points of intersection of the curve with $\text{x-}$axis
(ii) Find out the corresponding values of $\text{y}$ by putting \[\text{x}=\text{0}\] in the equation of the given curve, to estimate the points of intersection of the curve with $\text{y-}$axis
Step 4: Asymptotes
Find out the asymptotes of the curve.
(i) The vertical asymptotes of the given algebraic curve, or asymptotes parallel to the $\text{y-}$axis, are derived by equating the coefficient of the highest power of $\text{y}$ in the equation of the supplied curve to zero.
(ii) The horizontal asymptotes of the given algebraic curve, or asymptotes parallel to the $\text{x-}$axis, are derived by equating the coefficient of the highest power of $\text{x}$ in the equation of the supplied curve to zero.
Step 5: Region
Find out the regions of the plane in which no part of the curve lies. To determine such regions we solve the given equation for $\text{y}$ in terms of $\text{x}$ or vice-versa. Suppose that y becomes imaginary for\[\text{x}>\text{a}\], the curve does not lie in the region\[\text{x}>\text{a}\].
Step 6: Critical Points
Find out the values of $\text{x}$ at which $\dfrac{\text{dy}}{\text{dx}}=0$
At such points $\text{y}$ generally changes its character from an increasing function of $\text{x}$to a decreasing function of $\text{x}$ or vice-versa.
Step 7:
Trace the curve with the help of the above points.
Solved Examples
Definite Integration
Example - 1
Evaluate the Following Integrals:
(i) $\int\limits_{\text{2}}^{\text{3}}{{{\text{x}}^{\text{2}}}\text{dx}}$
(ii) $\int\limits_{\text{1}}^{\text{3}}{\dfrac{\text{x}}{\left( \text{x}+\text{1} \right)\left( \text{x}+\text{2} \right)}}\text{dx}$
Ans:
(i) $\int\limits_{\text{2}}^{\text{3}}{{{\text{x}}^{\text{2}}}\text{dx}}$
$=\left[ \dfrac{{{\text{x}}^{\text{3}}}}{\text{3}} \right]_{\text{2}}^{\text{3}}$
$=\dfrac{27}{3}-\dfrac{8}{3}$
$=\dfrac{19}{3}$
(ii) $\dfrac{\text{x}}{\left( \text{x}+\text{1} \right)\left( \text{x}+\text{2} \right)}=\dfrac{-\text{1}}{\text{x}+\text{1}}+\dfrac{\text{2}}{\text{x}+\text{2}}$ [Partial Fractions]
$\int\limits_{\text{1}}^{\text{3}}{\dfrac{\text{x}}{\left( \text{x}+\text{1} \right)\left( \text{x}+\text{2} \right)}\text{dx}}$
$=\left[ -\log \left| \text{x}+1 \right|+2\log \left| \text{x}+2 \right| \right]_{1}^{3}$
$=\left[ -\log \left| 4 \right|+2\log \left| 5 \right| \right]-\left[ -\log \left| 2 \right|+2\log \left| 3 \right| \right]$
$=\left[ -\log 4+2\log 5 \right]-\left[ -\log 2+2\log 3 \right]$
$=-2\log 2+2\log 5+\log 2-2\log 3$
$=-\log 2+\log 25-\log 9=\log 25-\log 18$
$=\log \dfrac{25}{18}$
Example - 2
Evaluate: $\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{secx}\text{.}\sqrt{\dfrac{\text{1}-\text{sinx}}{\text{1}+\text{sinx}}}}\text{dx}$
Ans:
$\text{I}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{secx}\text{.}\sqrt{\dfrac{\text{1}-\text{sinx}}{\text{1}+\text{sinx}}}}\text{dx}$
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{secx}\text{.}\sqrt{\dfrac{\text{1}-\text{sinx}}{\text{1}+\text{sinx}}}.\sqrt{\dfrac{\text{1}-\text{sinx}}{\text{1}+\text{sinx}}}}\text{dx}$
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{secx}\dfrac{\text{1}-\text{sinx}}{\sqrt{\text{1}-\text{si}{{\text{n}}^{\text{2}}}\text{x}}}}\text{dx}$
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{secx}\dfrac{\text{1}-\text{sinx}}{\text{cosx}}}\text{dx}$
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\left( \text{se}{{\text{c}}^{\text{2}}}\text{x}-\text{secxtanx} \right)\text{dx}}$
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{se}{{\text{c}}^{\text{2}}}\text{xdx}-\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{secxtanxdx}}}$
$=\left[ \text{tanx} \right]_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}-\left[ \text{secx} \right]_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}$
$=\left( \text{tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}-\text{tan0} \right)-\left( \text{sec}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}-\text{sec0} \right)$
$=\left( 1-0 \right)-\left( \sqrt{2}-1 \right)=2-\sqrt{2}$.
Example - 3
Evaluate: $\int\limits_{\text{-1}}^{\text{1}}{\text{5}{{\text{x}}^{\text{4}}}\sqrt{{{\text{x}}^{\text{5}}}\text{+1}}\text{dx}}$
Ans:
Let $\text{I}=\int\limits_{\text{-1}}^{\text{1}}{\text{5}{{\text{x}}^{\text{4}}}\sqrt{{{\text{x}}^{\text{5}}}\text{+1}}\text{dx}}$
Put ${{\text{x}}^{\text{5}}}=\text{t}$ so that $\text{5}{{\text{x}}^{\text{4}}}\text{dx}=\text{dt}$.
When $\text{x}=-1,\,\text{t}=-1$. When $\text{x}=1,\,\text{t}=1$.
$\text{I}=\int\limits_{\text{-1}}^{\text{1}}{\sqrt{\text{t+1}}\text{dt}}$
$=\left[ \dfrac{{{\left( \text{t}+\text{1} \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]_{-1}^{1}=\dfrac{2}{3}\left[ {{\left( \text{t}+\text{1} \right)}^{\dfrac{3}{2}}} \right]_{-1}^{1}$
$=\dfrac{2}{3}\left[ {{2}^{\dfrac{3}{2}}}-0 \right]=\dfrac{4\sqrt{2}}{3}$.
Example - 4
Prove that $\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\sqrt{\text{sin}\phi }}\text{co}{{\text{s}}^{\text{5}}}\phi \,\text{d}\phi =\dfrac{\text{64}}{\text{231}}$
Ans:
$\text{I}=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\sqrt{\text{sin}\phi }}\text{co}{{\text{s}}^{\text{5}}}\phi \,\text{d}\phi $
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\sqrt{\text{sin}\phi }}\text{co}{{\text{s}}^{4}}\phi \cos \phi \,\text{d}\phi $
$=\int\limits_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\sqrt{\text{sin}\phi }}{{\left( \text{1}-\text{si}{{\text{n}}^{\text{2}}}\phi \right)}^{\text{2}}}\text{cos}\phi \,\text{d}\phi $
Put $\sin \phi =\text{t}$ so that $\cos \phi \,\text{d}\phi =\text{dt}$.
When $\phi =0,\,\sin 0=\text{t}\Rightarrow \text{t=0}$
When $\phi =\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}},\,\text{sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}=\text{t}\Rightarrow \text{t}=\text{1}$
$\text{I}=\int\limits_{\text{0}}^{\text{1}}{\sqrt{\text{t}}}{{\left( \text{1}-{{\text{t}}^{\text{2}}} \right)}^{\text{2}}}\text{dt}=\int\limits_{\text{0}}^{\text{1}}{\sqrt{\text{t}}\left( -\text{2}{{\text{t}}^{\text{2}}}+{{\text{t}}^{\text{4}}} \right)\text{dt}}$
$=\int\limits_{0}^{1}{\left( {{\text{t}}^{\dfrac{\text{1}}{\text{2}}}}-\text{2}{{\text{t}}^{\dfrac{\text{5}}{\text{2}}}}+{{\text{t}}^{\dfrac{\text{9}}{\text{2}}}} \right)}\text{dt}$
$=\left[ \dfrac{{{\text{t}}^{\dfrac{\text{3}}{\text{2}}}}}{\dfrac{\text{3}}{\text{2}}}-\text{2}\dfrac{{{\text{t}}^{\dfrac{\text{7}}{\text{2}}}}}{\dfrac{\text{7}}{\text{2}}}+\dfrac{{{\text{t}}^{\dfrac{\text{11}}{\text{2}}}}}{\dfrac{\text{11}}{\text{2}}} \right]_{\text{0}}^{\text{1}}$
$=\left[ \dfrac{\text{2}}{\text{3}}{{\text{t}}^{\dfrac{\text{3}}{\text{2}}}}-\dfrac{\text{4}}{\text{7}}{{\text{t}}^{\dfrac{\text{7}}{\text{2}}}}+\dfrac{\text{2}}{\text{11}}{{\text{t}}^{\dfrac{\text{11}}{\text{2}}}} \right]_{\text{0}}^{\text{1}}$
$=\left[ \dfrac{2}{3}\left( 1 \right)-\dfrac{4}{7}\left( 1 \right)+\dfrac{2}{11}\left( 1 \right) \right]-\left[ 0-0+0 \right]$
$=\dfrac{2}{3}-\dfrac{4}{7}+\dfrac{2}{11}$
$=\dfrac{154-132+42}{231}=\dfrac{64}{231}$
Example-5
Evaluate:$\int\limits_{\text{1}}^{\text{2}}{\left( \dfrac{\text{x}-\text{1}}{{{\text{x}}^{\text{2}}}} \right)}{{\text{e}}^{\text{x}}}\text{dx}$ or $\int\limits_{\text{1}}^{\text{2}}{{{\text{e}}^{\text{x}}}\left( \dfrac{\text{1}}{\text{x}}-\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right)}\text{dx}$
Ans:
$\int{\left( \dfrac{\text{x}-\text{1}}{{{\text{x}}^{\text{2}}}} \right){{\text{e}}^{\text{x}}}\text{dx}=\int{{{\text{e}}^{\text{x}}}\left( \dfrac{\text{1}}{\text{x}}-\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right)\text{dx}}}$
$=\int{\dfrac{\text{1}}{\text{x}}}\text{.}{{\text{e}}^{\text{x}}}\text{dx}-\int{\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}}}\text{.}{{\text{e}}^{\text{x}}}\text{dx}$
$=\dfrac{\text{1}}{\text{x}}\text{.}{{\text{e}}^{\text{x}}}-\int{\left( -\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right){{\text{e}}^{\text{x}}}\text{dx}-\int{\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}}\text{.}{{\text{e}}^{\text{x}}}\text{dx}}}$ [Integrating first integral by parts]
$=\dfrac{\text{1}}{\text{x}}\text{.}{{\text{e}}^{\text{x}}}=\text{F}\left( \text{x} \right)$
$\int\limits_{\text{1}}^{\text{2}}{\left( \dfrac{\text{x}-\text{1}}{{{\text{x}}^{\text{2}}}} \right){{\text{e}}^{\text{x}}}\text{dx}=\left[ \dfrac{{{\text{e}}^{\text{x}}}}{\text{x}} \right]}_{\text{1}}^{\text{2}}$
$=\dfrac{\text{1}}{\text{2}}\text{.}{{\text{e}}^{\text{2}}}-\dfrac{\text{1}}{\text{1}}{{\text{e}}^{\text{1}}}=\dfrac{\text{1}}{\text{2}}{{\text{e}}^{\text{2}}}-\text{e}$.
Area Under the Curves
Example - 1
Find the area bounded by the curve $\text{y}={{\text{x}}^{\text{2}}}-\text{5x}+\text{6}$, $\text{X-}$axis and the lines $\text{x}=\text{1}$ and 4.
Ans:
For $\text{y}=\text{0}$, we get ${{\text{x}}^{\text{2}}}+5\text{x}+6=0\Rightarrow \text{x}=2,3$
Hence the curve crosses $\text{X-}$axis at $\text{x}=\text{2,3}$ in the interval $\left[ 1,4 \right]$.
Bounded Area$\text{=}\left| \int\limits_{\text{1}}^{\text{2}}{\text{ydx}} \right|\text{+}\left| \int\limits_{\text{2}}^{\text{3}}{\text{ydx}} \right|\text{+}\left| \int\limits_{\text{3}}^{\text{4}}{\text{ydx}} \right|$
$\Rightarrow \text{A=}\left| \int\limits_{\text{1}}^{\text{2}}{\left( {{\text{x}}^{\text{2}}}-\text{5x+6} \right)\text{dx}} \right|\text{+}\left| \int\limits_{\text{2}}^{\text{3}}{\left( {{\text{x}}^{\text{2}}}-\text{5x+6} \right)\text{dx}} \right|\text{+}\left| \int\limits_{\text{3}}^{\text{4}}{\left( {{\text{x}}^{\text{2}}}-\text{5x}+\text{6} \right)\text{dx}} \right|$
${{\text{A}}_{\text{1}}}=\left[ \dfrac{{{2}^{3}}-{{1}^{3}}}{3} \right]-5\left( \dfrac{{{2}^{2}}-{{1}^{2}}}{2} \right)+\left[ 6\left( 2-1 \right) \right]=\dfrac{5}{6}$
${{\text{A}}_{\text{2}}}=\dfrac{{{3}^{3}}-{{2}^{3}}}{3}-5\left( \dfrac{{{3}^{2}}-{{2}^{2}}}{2} \right)+6\left( 3-2 \right)=-\dfrac{1}{6}$
${{\text{A}}_{\text{3}}}=\dfrac{{{4}^{3}}-{{3}^{3}}}{3}-5\left( \dfrac{{{4}^{2}}-{{3}^{2}}}{2} \right)+6\left( 4-3 \right)=\dfrac{5}{6}$
$\Rightarrow \text{A}=\dfrac{5}{6}+\left| -\dfrac{1}{6} \right|+\dfrac{5}{6}=\dfrac{11}{6}\,\text{sq}\text{.}\,\text{units}$
Example - 2
Find the area bounded by the curve: $\text{y}=\sqrt{\text{4}-\text{x}}\text{,}$ $\text{X-}$axis and $\text{Y-}$axis.
Ans:
Trace the curve $\text{y}=\sqrt{\text{4}-\text{x}}$.
Put $\text{y}=\text{0}$ in the given curve to get $\text{x}=\text{4}$ as the point of intersection with X-axis.
Put $\text{x}=\text{0}$ in the given curve to get $\text{y}=2$ as the point of intersection with Y-axis.
For the curve, $\text{y}=\sqrt{\text{4}-\text{x}}\text{,}\,\text{4}-\text{x}\ge \text{0}$
$\Rightarrow \text{x}\le \text{4}$
$\Rightarrow $ curve lies only to the left of $\text{x}=4$ line.
As any $\text{y}$ is positive, curve is above X-axis.
Using step $1$ to $3$, we can draw the rough sketch of $\text{y}=\sqrt{\text{4}-\text{x}}$
In figure, Bounded area$=\int\limits_{\text{0}}^{\text{4}}{\sqrt{\text{4}-\text{x}}\text{dx}=\left| \dfrac{-\text{2}}{\text{3}}\left( \text{4}-\text{x} \right)\sqrt{\text{4}-\text{x}} \right|}_{\text{0}}^{\text{4}}=\dfrac{\text{16}}{\text{3}}\,\text{sq}\text{.}\,\text{units}$
Example - 3
$\text{AOBA}$ is the part of the ellipse $\text{9}{{\text{x}}^{\text{2}}}+{{\text{y}}^{\text{2}}}=\text{36}$ in the first quadrant such that $\text{OA}=\text{2}$ and \[\text{OB}=6\]. Find the area between the arc $\text{AB}$ and the chord $\text{AB}$.
Ans:
The given equation of the ellipse can be written as
$\dfrac{{{\text{x}}^{\text{2}}}}{\text{4}}+\dfrac{{{\text{y}}^{\text{2}}}}{\text{36}}=1\,\text{i}\text{.e}\text{.}\,\dfrac{{{\text{x}}^{\text{2}}}}{{{\text{2}}^{\text{2}}}}+\dfrac{{{\text{y}}^{\text{2}}}}{{{\text{6}}^{\text{2}}}}=1$
$\text{A}$ is $\left( 2,0 \right)$ and $\text{B}$ is $\left( 0,6 \right)$.
The equation of chord $\text{AB}$ is:
$\text{y}-0=\dfrac{6-0}{0-2}\left( \text{x}-\text{2} \right)$
$\Rightarrow \text{y}=-3\text{x}+6$.
Reqd. area (shown shaded)
$=\int\limits_{0}^{2}{3\sqrt{4-{{\text{x}}^{\text{2}}}}\text{dx}-\int\limits_{0}^{2}{\left( 6-3\text{x} \right)}}\text{dx}$
$=\text{3}\left[ \dfrac{\text{x}\sqrt{\text{4}-{{\text{x}}^{\text{2}}}}}{\text{2}}+\dfrac{\text{4}}{\text{2}}\text{si}{{\text{n}}^{-\text{1}}}\dfrac{\text{x}}{\text{2}} \right]_{\text{0}}^{\text{2}}-\left[ \text{6x}-\dfrac{\text{3}{{\text{x}}^{\text{2}}}}{\text{2}} \right]_{\text{0}}^{\text{2}}$
$=3\left[ \dfrac{2}{2}\left( 0 \right)+2{{\sin }^{-1}}\left( 1 \right) \right]-\left[ 6\left( 2 \right)-\dfrac{3\left( 4 \right)}{2} \right]$
$=3\left[ 2\times \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right]-\left[ 12-6 \right]$
$=\left( \text{3 }\!\!\pi\!\!\text{ }-\text{6} \right)\,\text{sq}\text{.}\,\text{units}$
Example - 4
Find the area bounded by the curves $\text{y}={{\text{x}}^{\text{2}}}$ and ${{\text{x}}^{\text{2}}}+{{\text{y}}^{\text{2}}}=2$ above X-axis.
Ans:
Let us first find the points of intersection of curves.
Solving $\text{y}={{\text{x}}^{\text{2}}}$ and ${{\text{x}}^{\text{2}}}+{{\text{y}}^{\text{2}}}=2$ simultaneously, we get:
${{\text{x}}^{\text{2}}}+{{\text{x}}^{\text{4}}}=2$
$\Rightarrow \left( {{\text{x}}^{\text{2}}}-1 \right)\left( {{\text{x}}^{\text{2}}}+2 \right)=0$
$\Rightarrow {{\text{x}}^{\text{2}}}=\text{1}\,\,\text{and}\,\,{{\text{x}}^{\text{2}}}=-\text{2}$ [reject]
$\Rightarrow \text{x}=\pm 1$
$\Rightarrow \text{A}=\left( -\text{1,1} \right)$ and $\text{B}=\left( 1,1 \right)$
Shaded Area$=\int\limits_{-\text{1}}^{\text{+1}}{\left( \sqrt{\text{2}-{{\text{x}}^{\text{2}}}}-{{\text{x}}^{\text{2}}} \right)\text{dx}}$
$=\int\limits_{-1}^{+1}{\sqrt{\text{2}-{{\text{x}}^{\text{2}}}}\text{dx}-\int\limits_{\text{-1}}^{\text{+1}}{{{\text{x}}^{\text{2}}}\text{dx}}}$
$=2\int\limits_{0}^{1}{\sqrt{\text{2}-{{\text{x}}^{\text{2}}}}\text{dx}-2\int\limits_{\text{0}}^{\text{1}}{{{\text{x}}^{\text{2}}}\text{dx}}}$
$=\text{2}\left[ \dfrac{\text{x}}{\text{2}}\sqrt{\text{2}-{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{2}}{\text{2}}\text{si}{{\text{n}}^{-\text{1}}}\dfrac{\text{x}}{\sqrt{\text{2}}} \right]_{\text{0}}^{\text{1}}-\text{2}\left( \dfrac{\text{1}}{\text{3}} \right)$
$=\text{2}\left( \dfrac{\text{1}}{\text{2}}+\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)-\dfrac{\text{2}}{\text{3}}=\dfrac{\text{1}}{\text{3}}\text{+}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\,\,\text{sq}\text{.}\,\,\text{units}$
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