1. ## Integral Question

Hey guys I need help with theses two question. I don't know how to do them and I'm just stuck, Just losted

Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.

and

Compute the indefinite integral

2. Originally Posted by killasnake
Hey guys I need help with theses two question. I don't know how to do them and I'm just stuck, Just losted

and
1. Evaluate $\displaystyle \int_{-4}^4 \sqrt{16-x^2} dx$ in term of the area
of a region.

Well this is the area of the upper half of the circle $\displaystyle x^2+y^2=4^2$, so its area is

$\displaystyle \frac{\pi r^2}{2}=\frac{\pi 4^2}{2}=8\pi$

RonL

3. Originally Posted by killasnake
Hey guys I need help with theses two question. I don't know how to do them and I'm just stuck, Just losted

and
Find:

$\displaystyle \int 5 x^4 + 3 \sec(x) tan(x)\ dx$

Look at the integrand:

$\displaystyle 5 x^4 + 3 \sec(x) \tan(x)=5x^4 +3 \frac{\sin(x)}{(\cos(x))^2}= \frac{d}{dx} \left[ x^5 + 3\frac{1}{\cos(x)}\right]$

So by the fundamental theorem of calculus:

$\displaystyle \int 5 x^4 + 3 \sec(x) tan(x)\ dx = x^5 + 3\frac{1}{\cos(x)} + C$

RonL

4. Hello, killasnake!

$\displaystyle \int\left[5x^4 + 3\sec(x)\tan(x)\right]\,dx$
We have: .$\displaystyle \underbrace{\int5x^4\,dx}_\downarrow \;+ \;3\!\underbrace{\int\sec(x)\tan(x)\,dx}_\downarro w$
. . . . . . . .$\displaystyle = \;\;x^5 \quad\;\;+ \qquad\;\;3\sec(x) + C$