Question

calc help plz hurry

Answer 1

@Vocaloid

Answer 2

I have to show my owrk but I believe it is false

Answer 3

By u-subs, let \(u = x^{-4}\). then \(du = -4x^{-5}dx\) so \(dx = \dfrac{du}{-4x^{-5}}\).
We need to also change the limits of integration, so \[u(2) = 2^{-4}, u(1) = 1^{-4} = 1.\] Then
\[
\begin{align}\int_1^2 -2x^{-5}e^{x^{-4} } dx &= \int_1^{2^{-4}} -2x^{-5} e^u \frac{du}{-4x^{-5}} \\ &= \frac{1}{2} \int_1^{2^{-4}} e^u du \\ &= \frac{1}{2}\big[e^u\big]_1^{2^{-4}} \\ &= \frac{1}{2}(e^{2^{-4}} - e^1) \\
&\approx -0.826 \approx -0.83
\end{align}\] It is true if you count approximations as correct.