Evaluate the integral by interpreting it in terms of areas.

37. \(\displaystyle \int^0_{-3} (1+\sqrt{9-x^2})*dx\)

the answer is \(\displaystyle 3+\frac{9}{4}*\pi\)

Where .. the heck did \(\displaystyle \pi\) come from?

I was trying to do

\(\displaystyle \frac{b-a}{n}\)

\(\displaystyle = \frac{-3-0}{n}\)

\(\displaystyle =\frac{-3}{n}\)

and then \(\displaystyle x_{i}\) as \(\displaystyle \frac{-3i}{n}-3\)

then plugging \(\displaystyle x_{i}\) into \(\displaystyle f(x)*dx\) then solving for the limit. It came out to be a super complicated root problem so I look in the back to see if I'm in the right track.. and theres a \(\displaystyle \pi\) in the answer...

I think I'm missing something here... how is this integral end up having a \(\displaystyle \pi\)?

The key to this problem is "Evaluate the integral by interpreting it in terms of

**areas**.

\(\displaystyle \int^0_{-3} (1+\sqrt{9-x^2})*dx= \int_{-3}^0 1 dx+ \int_{-3}^0 \sqrt{9- x^2}dx\)

We can interpret the first integral as the area under the line y= 1 between x= -3 and 0. That is the area of a rectangle with width 3 and height 1.

Squaring both sides of \(\displaystyle y= \sqrt{9- x^2}\) gives \(\displaystyle y^2= 9- x^2\) or \(\displaystyle x^2+ y^2= 9\), a circle with center at (0, 0) and radius 3. The square root is the upper half of that and from -3 to 0 gives the upper left quadrant. That integral is 1/4 the area of circle of radius 3. (That's where the "\(\displaystyle \pi\)" comes in.)