Question

How do you evaluate the integral of $\int{(3-2x)dx}$ from -1 to 3?

Answer 1

Hint: To evaluate the integral of a polynomial, we need to consider the inverse of the power rule used in differentiation, i.e., $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$ . If the integral value of the function is within the given bounds, we subtract the value of the function at the lower bound from the function at the higher bound.

Complete Step by Step Solution:
Considering the lower and upper limits the question can be written as
$=\int\limits_{-1}^{3}{(3-2x)dx}$
Expanding the given integral, we get
$=\int\limits_{-1}^{3}{3dx-\int\limits_{-1}^{3}{2xdx}}$
Integrating the above integral,
$=\left. 3x-2\left( \dfrac{{{x}^{2}}}{2} \right) \right|_{-1}^{3}$
Simplifying,
$=\left. 3x-\left( {{x}^{2}} \right) \right|_{-1}^{3}$
Applying the bounds and substituting x values,
$=\left[ 3(3)-{{(3)}^{2}} \right]-\left[ 3(-1)-{{(-1)}^{2}} \right]=\left[ 9-9 \right]-\left[ -3-1 \right]$
Simplifying the above expression, we get
$=[0]-[-4]=4$

Therefore, $\int\limits_{-1}^{3}{(3-2x)dx}=4$

Note:
Alternate Method:
Taking the lower and upper limits into consideration the question can be written as
$=\int\limits_{-1}^{3}{(3-2x)dx}$
Using the formula, $\int\limits_{a}^{b}{f(x)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}$
In the above formula, for each positive integer n, we have $\Delta x=\dfrac{b-a}{n}$ and for $i=1,2,3,.......,n,$ we get ${{x}_{i}}=a+i\Delta x$ .
For every n, $\Delta x=\dfrac{b-a}{n}=\dfrac{3-(-1)}{n}=\dfrac{4}{n}$
For ${{x}_{i}}=a+i\Delta x$ we get the expression as ${{x}_{i}}=-1+i\dfrac{4}{n}=-1+\dfrac{4i}{n}$
Using this, we can write the expression of $f\left( {{x}_{i}} \right)$ as $f\left( {{x}_{i}} \right)=3-2{{x}_{i}}=3-2\left( -1+\dfrac{4i}{n} \right)=5-\dfrac{8i}{n}$
Using summation formulas for simplifying the above expression,
$\Rightarrow \sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}=\sum\limits_{i=1}^{n}{\left( 5-\dfrac{8i}{n} \right)\dfrac{4}{n}}$
Multiplying each term in the bracket with $\dfrac{4}{n}$
$=\sum\limits_{i=1}^{n}{\left( \dfrac{20}{n}-\dfrac{32i}{{{n}^{2}}} \right)}$
$=\dfrac{20}{n}\sum\limits_{i=1}^{n}{\left( 1 \right)+\dfrac{32}{{{n}^{2}}}}\sum\limits_{i=1}^{n}{\left( i \right)}$
Simplifying the summation in each of the terms as $\sum\limits_{i=1}^{n}{\left( 1 \right)=n}$ and $\sum\limits_{i=1}^{n}{\left( i \right)}=\dfrac{n(n+1)}{2}$ we can simplify as $=\dfrac{20}{n}[n]-\dfrac{32}{{{n}^{2}}}\left[ \dfrac{n(n+1)}{2} \right]$
So, $\sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}=20-\dfrac{32}{2}\left[ \dfrac{n(n+1)}{{{n}^{2}}} \right]$
To solve the expression, we need to evaluate the limit as $n\to \infty $ . This implies that we need to evaluate $\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{n(n+1)}{{{n}^{2}}} \right]$. The numerator of this expression will expand to a polynomial which has a leading term ${{n}^{2}}$, therefore, the limit as $n\to \infty $ is 1.
$\Rightarrow \dfrac{n(n+1)}{{{n}^{2}}}=\left( \dfrac{n}{n} \right)\left( \dfrac{n+1}{n} \right)=(1)\left( \dfrac{n}{n}+\dfrac{1}{n} \right)=1+\dfrac{1}{n}$
Simplifying and completing the integration, we get
$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}=20-\dfrac{32}{2}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{n(n+1)}{{{n}^{2}}} \right]$
Applying the limit and simplifying,
$\Rightarrow 20-16=4$
Therefore, $\int\limits_{-1}^{3}{(3-2x)dx}=4$