MHB Calculating Area Under a Curve: Is My Approach Correct?

[jaychay]

Can you please check it for me that I have done it wrong or not ?
Thank you in advance.

 

[skeeter]

w/ respect to x, top function - bottom function

$\displaystyle \int_{-1}^0 9 - 9^{-x} \, dx + \int_0^2 9 - 3^x \, dx$

integral w/respect to y is ok and it can be simplified ...

$\displaystyle \dfrac{3}{2\ln{3}} \int_1^9 \ln{y} \, dy$

secant function integral is set up correctly ... antiderivative is rather easy to see if you recognize the chain rule

 

[HOI]

The second problem looks pretty standard- take x from 0 to $\sqrt{\frac{\pi}{3}}$ and, for each x, y from 0 to $2xsec^2(x^2)$.

The area is $\int_0^{\sqrt{\frac{\pi}{3}}} 2x sec^2(x^2)dx$.

Let $u= x^2$ so that $du= 2xdx$. When x= 0, u=0 and when x= $\sqrt{\frac{\pi}{3}}$, $u= \frac{\pi}{3}$.

The integral becomes $\int_0^{\frac{\pi}{3}} sec^2(u)du$.