Please check my answers: find area bounded by y=x^2, 4x-3x^2

[XBOX999]

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area S of the region. (Give an exact answer.)

1) y=x^2 ,4x-3x^2
2) 3x+y^2=13 , x= 4y

My answers
1)I have got 8/3

2) I have got 67/3

Please check it to me and tell me the right answers.

 

[Deleted member 4993]

Re: Please I want you to check my answers to this quetion

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area S of the region. (Give an exact answer.)

1) y=x^2 ,4x-3x^2
2) 3x+y^2=13 , x= 4y


My answers
1)I have got 8/3 <<< I get different answer. If you want us to check your work - you need to show your work.


2) I have got 67/3<<< I get different answer. If you want us to check your work - you need to show your work.


Please check it to me and tell me the right answers.

 

[BigGlenntheHeavy]

Don't understand # 1.

For # 2, by integrating with respect to y, we need only one integral.

Let x = g(y) = (13-y²)/3 and let x = f(y) = 4y. These two curves intersect when y = -13 and y = 1. Since f(y)?g(y) on this interval, we have:

\(\displaystyle \Delta A = [g(y)-f(y)]\Delta y = [{\frac{13-y^{2}}{3}-4y]\Delta y.\)

\(\displaystyle Hence, \ the \ area \ is: \ \int_{-13}^{1}[\frac{13-y^{2}}{3}-4y]dy = \frac{1372}{9}.\)

 

[BigGlenntheHeavy]

OK, for # 1, I assume you have y = x^2, y = 4x-3x^2.

\(\displaystyle Then \ A = \int_{0}^{1}[(4x-3x^{2})-x^{2}]dx = \frac{2}{3}.\)