MHB Area of a Bounded Region: Calculation Examples

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The discussion focuses on verifying the calculations for the areas of bounded regions defined by specific curves. The area under the curve y=9-x^2 and the x-axis is confirmed to be 36 square units, with a suggestion to simplify calculations using the even-function rule. For the area bounded by ln(x) from x=2 to x=4, the correct answer is noted as approximately 2.159 square units, with an exact expression provided. The area under sin(x) from x=(1/3)pi to x=(2/3)pi is confirmed to be 1 square unit. Participants are encouraged to limit questions to two per topic and consider using computational tools for verification.
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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.

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

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

my answer is 36 sq. units.

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

soln.

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

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

 
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1.) Correct. You could use the even-function rule to make your calculation a bit simpler:

$$A=2\int_0^3 9-x^2\,dx$$

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

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

or

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

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)
 
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