FUNDAMENTAL THEOREM OF CALCULUS (PART-1)

If $f$(x) is a continuous function on $\left[a,b\right],$ then the function $\mathrm{g(x)}$ defined by

$$g\left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)dt,a\le x\le b$$

is an antiderivative of $f$, that is

$$g^{\prime }\left(x\right)=f\left(x\right)\text{or}\frac{d}{dx}\left(\underset{a}{\overset{x}{\int }}f\left(t\right)dt\right)=f\left(x\right).$$

Know about fundamental theorem of calculus Part 2 click here.

EXAMPLE:

Find the derivative of the function $f\left(x\right)=\underset{0}{\overset{x^2}{\int }}\sqrt{1+t^2}dt$

Solution:

Since the upper limit of integration is not $x,$ we apply the chain rule. Let $u=x^2,$ then $u^{\prime }=2x.$

Consider the new function

$$h\left(u\right)=\underset{0}{\overset{u}{\int }}\sqrt{1+t^2}dt.$$

By the FTC1, we can write

$$h^{\prime }\left(u\right)=\sqrt{1+u^2}.$$

As $f\left(x\right)=h\left(x^2\right),$ we have

$$f^{\prime }\left(x\right)={\left[h\left(x^2\right)\right]}^{\prime }=h^{\prime }\left(x^2\right)\cdot {\left(x^2\right)}^{\prime }=\sqrt{1+{\left(x^2\right)}^2}\cdot 2x=2x\sqrt{1+x^4}.$$

HOMEWORK:

1.  Find the derivative of the function 

$g\left(x\right)=\underset{3}{\overset{x^2}{\int }}\frac{dt}{t}.$