Differentiating an Integral then Calculating?

Question

I've been having an issue with an integral where I have to plug in a number after doing the derivative of it. The problem is this:

G'(2), where G(x) = $\int_{0}^{x^3} \sqrt{t+8 } dt$

I tried integrating it and got:

$\sqrt{x^3+8} * 3x^2$

Am I forgetting to plug something in after?

Answer 1

$\int_0^{x^3}(t+8)^{-\frac{1}{2}}dt$ integrates to

$[\frac{2}{3}(t+8)^{\frac{3}{2}}]_0^{x^3}$

which gives you

$G(x)=\frac{2}{3}(x^3+8)^{\frac{3}{2}}-\frac{2}{3}(0+8)^{\frac{3}{2}}$

differentiating gives you

$G'(x)=(x^3+8)^{\frac{1}{2}}.3x^2$

so

$G'(2)=(8+8)^{\frac{1}{2}}.12=48$

Answer 2

You applied the fundamental theorem of calculus and the chain rule to deduce that $$ G'(x)=\sqrt{x^3 +8}\times 3x^2 $$ The question asks for $G'(2)$ i.e. $$ G'(2)=\sqrt{2^3 +8}\times 3(2^2) $$

Answer 3

For $x>0$ the integral is:

$$-\frac{2}{3} \left(-\sqrt{x^3+8} x^3-8 \sqrt{x^3+8}+16 \sqrt{2}\right)$$

The derivative is then

$$\frac{2}{3} \left(-\frac{3 x^5}{2 \sqrt{x^3+8}}-3 \sqrt{x^3+8} x^2-\frac{12 x^2}{\sqrt{x^3+8}}\right)$$

Just plug in $x=2$ to get $-48$.