Author Topic: Integration/Area under curve help (Read 1694 times) Tweet Share

Rikahs · « **on:** October 17, 2016, 07:35:49 pm »

Hi,

How would I answer the attached question? For some reason Im getting a negative answer.

Blissfulmelodii · « **Reply #1 on:** October 17, 2016, 07:51:16 pm »

Quote from: Rikahs on October 17, 2016, 07:35:49 pm

Hi,

How would I answer the attached question? For some reason Im getting a negative answer.

Umm not sure if this is right but I would possibly find the area of the quarter circle and then minus from that the area under the curve between the circle and the parabola using the y-axis boundaries... if that makes sense.

RuiAce · « **Reply #2 on:** October 17, 2016, 07:57:03 pm »

Quote from: Rikahs on October 17, 2016, 07:35:49 pm

Hi,

How would I answer the attached question? For some reason Im getting a negative answer.

$\text{Blissfulmelodii is correct.}\\ \text{Note that the area of the quadrant is }\frac{1}{4}\times \pi r^2 = \pi$
$\text{Observe that the area of the quadrant is too much.}\\ \text{We need to eliminate that area under the parabola between }x=0\text{ and }x=1$
$\text{So the actual answer is}\\ A=\pi - \int_0^1(x^2-3x+2)dx$

Rikahs · « **Reply #3 on:** October 17, 2016, 08:55:41 pm »

Thank you both!!! For some reason I didn't see that.

Blissfulmelodii · « **Reply #4 on:** October 17, 2016, 08:57:36 pm »

Quote from: Rikahs on October 17, 2016, 08:55:41 pm

Thank you both!!! For some reason I didn't see that.

No worries! I'm just glad I could actually answer a math question haha

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Author Topic: Integration/Area under curve help (Read 1694 times) Tweet Share

Rikahs

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