# Thread: one of those "stupid" proofs

1. ## one of those "stupid" proofs

Hi all, I have to do this problem according to the first part of the Fundamental Theorem of Calculus. Fundamental theorem of calculus - Wikipedia, the free encyclopedia

h(x)= (integral of 0 to x^2) sqrt(1+r^3) dr

f(x)=sqrt(1+r^3)

h'(x)=f(x)

h'(x)=sqrt(1+x^6)* 2x

Am I right? I was kind of confused, also did I miss any steps? Thanks!

2. it's correct.

3. Originally Posted by soyeahiknow
Hi all, I have to do this problem according to the first part of the Fundamental Theorem of Calculus. Fundamental theorem of calculus - Wikipedia, the free encyclopedia

h(x)= (integral of 0 to x^2) sqrt(1+r^3) dr

f(x)=sqrt(1+r^3)

h'(x)=f(x)

h'(x)=sqrt(1+x^6)* 2x

Am I right? I was kind of confused, also did I miss any steps? Thanks!
Dear soyeahiknow,

Another approach is,

$h(x)=\int^{x^2}_{0}\sqrt{1+r^3}~dr$

Take $y=x^2\Rightarrow{\frac{dy}{dx}=2x}$

$\frac{d}{dy}h(x)=\frac{d}{dy}\int^{y}_{0}\sqrt{1+r ^3}~dr$

$\frac{d}{dx}h(x)\times{\frac{dx}{dy}}=\sqrt{1+y^3}$

$h'(x)=\sqrt{1+y^3}\times\frac{dy}{dx}=\sqrt{1+x^6} \times{2x}$

Hope this helps.