# find f(2) given the function defined by $\int _0^{x^2} f(t)= x^2(x+1)$

$f(t)$ is continuous and satisfies the condition for x>=0 $$\int _0^{x^2} f(t)\,\mathrm{d}t= x^2(x+1)$$

how do I find f(2) plug the 2 in the right side, or do so multiplying by $x^2$ derivative?

• I added the variable of integration to the integral. Check it to make sure I am understanding what you are asking.
– robjohn
Commented Jun 22, 2017 at 1:08
• It is correct, my mistake
– Jrgs
Commented Jun 22, 2017 at 1:30

Differentiating both sides you get $$f(x^2) 2x=2x(x+1)+x^2$$

Plug in $x=\sqrt{2}$.

• So if have another f(x) in the integration limit i shall find the derivative and plug an x that makes f(x)=f(2)?
– Jrgs
Commented Jun 22, 2017 at 0:19

Integrate: $$F(x^2)-F(0)=x^3+x^2$$ Change: $$x^2=t \Rightarrow x^3=t^{3/2}$$ Get: $$F(t)-F(0)=t^{3/2}+t$$ Differentiate: $$f(t)=\frac{3}{2}t^{1/2}+1$$ Plug $t=2:$

$$f(2)=\frac{3}{2}\sqrt{2}+1.$$

By $t=u^2$,

$$2\int_a^xf (u^2)udu=\int_a^x (3u^2+2u) du$$

$$2 (x-a)f (c^2)c=(x-a)(3d^2+2d)$$

with $a\le c,d\le x$. take $a=\sqrt {2}$ and make $x\to a$.