4
$\begingroup$

Let $h(x)=\int_2^{10/x}arctan(t)dt$. Use the fundamental Theorem of Calculus to find $h'(x)$.

So I believe the answer is $arctan(10/x)-arctan(2)$, but am being told it is incorrect. Any other ideas?

My thought process was that the derivative of an integral will give back the integrand, then evaluate $F(b)-F(a)$ where F is the original integrand.

Thanks!

$\endgroup$
  • 1
    $\begingroup$ Apply the Fundamental Theorem of Calculus (and apply chain rule) $\endgroup$ – Morgan Rodgers Jan 24 at 0:43
2
$\begingroup$

if $F(x) - F(b) = \int_b^{x} f(t) \ dt$

$F'(x) = f(x)$

This is the fundamental theorem of calculus.

We have

$F(\frac {10}{x}) - F(b) = \int_b^{\frac{10}{x}} f(t) \ dt$

$\frac {d}{dx}\left(F(\frac {10}{x}) - F(b)\right) = \frac {d}{dx}\int_b^{\frac{10}{x}} f(t) \ dt$

Remember, b is a constant, so $\frac {d}{dx} F(b) = 0$

For the other term we are going to use the chain rule.

$F'(\frac {10}{x}) = f(\frac {10}{x})(\frac {d}{dx} \frac {10}{x}) = f(\frac{10}{x})(\frac {-10}{x^2})$

$-\frac {10\arctan \frac {10}{x}}{x^2}$

$\endgroup$
3
$\begingroup$

It's true that if $F(x)=\int_a^x f(t)\,dt$, then $F'(x)=f(x)$. But in your case you have $$ h(x)=F(\tfrac{10}x), $$ where $F(x)=\int_2^x\arctan(t)\,dt$. So to find $h'$ you need to apply the Chain Rule and the Fundamental Theorem of Calculus.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.