FUNDAMENTAL THEOREM OF CALCULUS (PART-2)

 FUNDAMENTAL THEOREM OF CALCULUS (PART-2)

The second part of the fundamental theorem tells us how we can calculate a definite integral.

If $\mathrm{f(x)}$ is a continuous function on $\left[a,b\right]$ and $\mathrm{F(x)}$ is an antiderivative of $\mathrm{f(x)},$ that is ${\mathrm{F\text{'}(x)=}}^{}f(x),$ then

$$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=F\left(b\right)-F\left(a\right)\text{or}\underset{a}{\overset{b}{\int }}{F}^{\prime }\left(x\right)dx=F\left(b\right)-F\left(a\right).$$

To evaluate the definite integral of a function $f$ from $a$ to $b,$ we just need to find its antiderivative $F$ and compute the difference between the values of the antiderivative at b and a.  

EXAMPLE:  

Evaluate the integral $\underset{0}{\overset{1}{\int }}\frac{x}{{\left(3x^2-1\right)}^4}dx$

Solution.

First we make the substitution:

$$t=3x^2-1,\Rightarrow dt=6xdx,\Rightarrow xdx=\frac{dt}{6}.$$

Determine the new limits of integration. When $x=0,$ then $t=-1.$ When $x=1,$ then we have $t=2.$ So, the integral with the new variable $t$ can be easily calculated:

$$\underset{0}{\overset{1}{\int }}\frac{x}{{\left(3x^2-1\right)}^4}dx=\underset{-1}{\overset{2}{\int }}\frac{\frac{dt}{6}}{t^4}=\frac{1}{6}\int t^{-4}dt=\frac{1}{6}{\left(\frac{t^{-3}}{-3}\right)|}_{-1}^2=-\frac{1}{18}\left(\frac{1}{8}-1\right)=\frac{7}{144}.$$

HOME WORK:- 

1. Evaluate the integral 

$\underset{0}{\overset{\ln 2}{\int }}x{e}^{-x}d\mathrm{x\; .}$