# evaluating integral

#### dorkymichelle

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
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.

#### dorkymichelle

how did the sin come into this?
i don't see how the original function is of a trig graph

#### matheagle

MHF Hall of Honor
how did the sin come into this?
i don't see how the original function is of a trig graph
This is a trig sub, make that substitution and use Pythagoras' theorem.

#### HallsofIvy

MHF Helper
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.)