evaluating integral

Dec 2009
103
0
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 \(\displaystyle \pi\)?
 

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
Write these as two integrals.
Integrating the 1 and plugging in your bounds will give you the 3.
Next integrate the second integral with the substitution of
\(\displaystyle x=3\sin u\) and see what you get.
 
Dec 2009
103
0
how did the sin come into this?
i don't see how the original function is of a trig graph
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.)