Fundamental Theorem of Calculus improper integral question

Question

I want to find the derivative of a function G(x) defined by:

$$ G(t) = \int_{0}^{t^2} x^{2} \sin(x^{2}) \, dx$$

Am I correct in evaluating the answer to be $= 2t^{3}\sin(t^{2})$?

What I did was basically substitute the upper limit into $x^{2}$ and multiplied the whole thing by the derivative of $t^{2}$. Can someone help me understand a general formula on how to calculate problems like these, especially if the upper and lower limits are less straightforward and when the function to be integrated is also more complicated?

Would appreciate any references to resources! Thank you very much and cheers!

Answer 1

Define a function $$F(x) = \int x^2\sin (x^2) dx$$

Then, by the Newton-Leibniz formula $$G(t) = \int_{0}^{t^2} x^{2} \cdot \sin(x^{2}) \cdot dx = F(t^2) - F(0)$$

Now, seeing that $F(0)$ is a constant, the derivative of $G(t)$ is given by $$G'(t) = \frac{d}{dt}F(t^2)$$

From the definition of $F(x)$ it follows that $F'(t)=t^2\sin t^2$, and hence $$\frac{d}{dt}F(t^2) = \frac{dF(t^2)}{d(t^2)} \cdot \frac{d(t^2)}{dt}$$

We conclude that $G'(t) = 2t^5 \sin (t^4)$.

Answer 2

No, that's not the right way to do it.

Consider instead $$ F(t)=\int_0^t x^2\sin(x^2)\,dx $$ Then you can write $G(t)=F(t^2)$, so $$ G'(t)=2tF'(t^2) $$ by the chain rule. Since $F'(t)=t^2\sin(t^2)$, you have $$ G'(t)=2t\cdot t^4\sin(t^4)=2t^5\sin(t^4) $$

Answer 3

Hint

IMHO, I would suggest to simply remember that the fundamental theorem of calculus write $$g(x)=\frac d {dx} \, \int_a^{b(x)} f(t)\,dt=f(b(x))\, b'(x)$$