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Hint:
Here we will show how to use Fundamental Theorem of Calculus. Firstly we will state the theorem of calculus. Then we will give an example and solve it using the statement of the theorem to show how to use the theorem. There are two categories of Fundamental Theorem of Calculus which is used to solve an integral function.
Complete step by step solution:
Fundamental Theorem of Calculus is divided into two parts:
First Fundamental Theorem of Calculus states that if a function $f$ is continuous on $\left[ {a,b} \right]$ then the function defined by
$S\left( x \right) = \int\limits_a^x {f\left( t \right)dt} $
Is continuous on $\left[ {a,b} \right]$ and differential on $\left( {a,b} \right)$and $S'\left( x \right) = f\left( x \right)$ which can be written as,
$\dfrac{d}{{dx}}\int\limits_a^x {f\left( t \right)dt = f\left( x \right)} $
Example- Find the derivative of $k\left( x \right) = \int\limits_2^x {\left( {{4^t} + t} \right)dt} $
As we can see the function inside the integral is continuous so comparing the example with the theorem we get,
$S\left( x \right) = k\left( x \right)$ And $f\left( x \right) = {4^t} + t$
Therefore,
$k'\left( x \right) = {4^x} + x$
Second Fundamental Theorem of Calculus states that integral of a function over some interval can be computed by using any one of the infinite anti-derivatives it has. So if $f$ is continuous on $\left[ {a,b} \right]$ then the function defined by
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$
Where, $F$ is an anti-derivative of$f$.
Example- Evaluate $\int\limits_1^2 {{x^4}dx} $
Comparing the above example with the theorem we get,
$f\left( x \right) = {x^4}$ And also
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$….$\left( 1 \right)$
As, $F$ is an anti-derivative of $f$ so on Integrating $f$ we get,
$F\left( x \right) = \dfrac{{{x^{4 + 1}}}}{{4 + 1}} = \dfrac{{{x^5}}}{5}$
Now, $a = 1,b = 2$ so value of anti-derivative is,
$F\left( 1 \right) = \dfrac{{{1^5}}}{5} = \dfrac{1}{5}$
$F\left( 2 \right) = \dfrac{{{2^5}}}{5} = \dfrac{{32}}{5}$
Substituting the above value in equation $\left( 1 \right)$ we get,
$\int\limits_a^b {f\left( x \right)dx} = \dfrac{{32}}{5} - \dfrac{1}{5}$
Taking L.C.M on right side we get,
$\Rightarrow \int\limits_a^b {f\left( x \right)dx} = \dfrac{{32 - 1}}{5} \\
\Rightarrow \int\limits_a^b {f\left( x \right)dx} = \dfrac{{31}}{5} \\ $
So, $\int\limits_1^2 {{x^4}dx} = \dfrac{{31}}{5}$
Note:
The Fundamental Theorem of Calculus is the main medium to study calculus. This is used to find the relation between the differentiation and integration. The first part of the theorem deals with the anti-derivative of a derivative while the second part deals with the relation between indefinite integrals and antiderivatives.
Here we will show how to use Fundamental Theorem of Calculus. Firstly we will state the theorem of calculus. Then we will give an example and solve it using the statement of the theorem to show how to use the theorem. There are two categories of Fundamental Theorem of Calculus which is used to solve an integral function.
Complete step by step solution:
Fundamental Theorem of Calculus is divided into two parts:
First Fundamental Theorem of Calculus states that if a function $f$ is continuous on $\left[ {a,b} \right]$ then the function defined by
$S\left( x \right) = \int\limits_a^x {f\left( t \right)dt} $
Is continuous on $\left[ {a,b} \right]$ and differential on $\left( {a,b} \right)$and $S'\left( x \right) = f\left( x \right)$ which can be written as,
$\dfrac{d}{{dx}}\int\limits_a^x {f\left( t \right)dt = f\left( x \right)} $
Example- Find the derivative of $k\left( x \right) = \int\limits_2^x {\left( {{4^t} + t} \right)dt} $
As we can see the function inside the integral is continuous so comparing the example with the theorem we get,
$S\left( x \right) = k\left( x \right)$ And $f\left( x \right) = {4^t} + t$
Therefore,
$k'\left( x \right) = {4^x} + x$
Second Fundamental Theorem of Calculus states that integral of a function over some interval can be computed by using any one of the infinite anti-derivatives it has. So if $f$ is continuous on $\left[ {a,b} \right]$ then the function defined by
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$
Where, $F$ is an anti-derivative of$f$.
Example- Evaluate $\int\limits_1^2 {{x^4}dx} $
Comparing the above example with the theorem we get,
$f\left( x \right) = {x^4}$ And also
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$….$\left( 1 \right)$
As, $F$ is an anti-derivative of $f$ so on Integrating $f$ we get,
$F\left( x \right) = \dfrac{{{x^{4 + 1}}}}{{4 + 1}} = \dfrac{{{x^5}}}{5}$
Now, $a = 1,b = 2$ so value of anti-derivative is,
$F\left( 1 \right) = \dfrac{{{1^5}}}{5} = \dfrac{1}{5}$
$F\left( 2 \right) = \dfrac{{{2^5}}}{5} = \dfrac{{32}}{5}$
Substituting the above value in equation $\left( 1 \right)$ we get,
$\int\limits_a^b {f\left( x \right)dx} = \dfrac{{32}}{5} - \dfrac{1}{5}$
Taking L.C.M on right side we get,
$\Rightarrow \int\limits_a^b {f\left( x \right)dx} = \dfrac{{32 - 1}}{5} \\
\Rightarrow \int\limits_a^b {f\left( x \right)dx} = \dfrac{{31}}{5} \\ $
So, $\int\limits_1^2 {{x^4}dx} = \dfrac{{31}}{5}$
Note:
The Fundamental Theorem of Calculus is the main medium to study calculus. This is used to find the relation between the differentiation and integration. The first part of the theorem deals with the anti-derivative of a derivative while the second part deals with the relation between indefinite integrals and antiderivatives.