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Area of a region.


Active member
May 13, 2013
just want to check if my solutions were correct..

1. find the area of the bounded region by the curve y=9-x^2 and the x-axis.

$\displaystyle 9-x^2=0$ the roots or points of intersection are x=3,-3

$\displaystyle\int_{-3}^3 (9-x^2)dx$ = $\displaystyle \left[9x-\frac{x^3}{3}\right]_{-3}^3$

my answer is 36 sq. units.

2. find the area bounded by ln(x) and x-axis with x=2 and x=4.


$\displaystyle \int_{2}^4 ln(x)dx=\left[xlnx-x\right]_{2}^4$

my answer is 2.158 sq. units

3. find the area bounded by sin(x), x-axis, x=(1/3)pi, x=(2/3)pi

$\displaystyle\int_{\frac{1}{3}\pi}^{\frac{2}{3} \pi}\sin(x)dx=[-\cos(x)]_{\frac{1}{3}\pi}^{\frac{2}{3} \pi}$

my answer is 1 sq. units



Staff member
Feb 24, 2012
1.) Correct. You could use the even-function rule to make your calculation a bit simpler:

\(\displaystyle A=2\int_0^3 9-x^2\,dx\)

2.) You have rounded incorrectly (to 3 decimal places it is 2.159). I would instead write the exact result as:

\(\displaystyle A=2\left(3\ln(2)-1 \right)\)


\(\displaystyle A=2\ln\left(\frac{8}{e} \right)\)

3.) Correct.

In the future, please limit a topic to no more than two questions, and if you are just trying to see if you have integrated correctly (assuming you have set up the correct integral to solve the given problem), then I recommend either Wolfram|Alpha: Computational Knowledge Engine or one of our widgets. :D

If you are instead curious if you have set up the integral correctly, then by all means we encourage you to ask here. (Sun)