Question

If `F(x)=int_1^(x^2)sqrt(t^2+3)dt`, then `F'(2)=`?

I anti-differentiated this function and plugged the value in, I got `sqrt(19)^3*8/3-16/3`. However, the correct answer is `4sqrt(19)`.

Is my answer wrong or I need to simplify it? If so, how?

Thank you!

Answer 1

We need to use the Chain Rule in conjunction with FToC part I. See below.

Explanation:

This question requires application of the first part of the Fundamental Theorem of Calculus, which tells us if

`F(x)=int_a^xf(t)dt` where `a` is any constant, then

`F'(x)=f(x)`

This makes sense, it connects our knowledge of differentiation and integration together, telling us that the derivative of an integrated function is really just that function itself.

Here, however, our definite integral involves `x^2` rather than `x.` This will require an application of the Chain Rule.

Let `u=x^2`. We then have

`F(x)=int_1^usqrt(t^2+3)dt`

We see `f(t)=sqrt(t^2+3)`, and `f(u)=sqrt(u^2+3)`

where `u^2=(x^2)^2=x^4`

Thus,

`F'(x)=sqrt(u^2+3)(du)/dx`

`(du)/dx=d/dx(x^2)=2x`

So,

`F'(x)=2xsqrt(x^4+3)`

Then,

`F'(2)=4sqrt(16+3)=4sqrt19`