#### To determine

**To evaluate:**

The integral

∫-301+9-x2dx

by interpreting it in terms of area

#### Answer

3+9π4

#### Explanation

**1) Concept:**

A definite integral can be interpreted as a net area, that is, as difference of the areas;

∫abf(x)dx=A1-A2

where A1 is the region above x- axis and below the graph of f, and A2 is the region below x- axis and above the graph of f.

**2) Formula:**

i. Area of rectangle : A=l·b

ii. Area of quarter circle = A=14·π·r2

**3) Given:**

∫-301+9-x2dx

**4) Calculation:**

The integral ∫-301+9-x2dx can be interpreted as the area under the graph of

fx= 1+9-x2 between x= -3 and x=0

The graph of 1+9-x2 in the given domain is shown below.

From the graph,A1 represents the area of the rectangle and A2 represents one quarter of the area of circle with radius r=3.

Therefore, the area of a rectangle is

A1=l·b=3·1=3

And thearea of aquarter circle is

A2=14·π·r2=14·π·32=9π4

Therefore, the net area is the addition of the two areas, since both are above the x axis

∫-301+9-x2dx=A1+A2=3+9π4

**Conclusion:**∫-301+9-x2dx=3+9π4