Integration of a Semi Circle

[rinachez]

I am trying to find the area of the semi circle using integration given this velocity [ f'(x) ] graph



I figured out the equation of the circle but I don't know how to integrate it. Would trig substitution work for this problem since it uses coord points or am I just approaching it incorrectly?
I know how to do the geometry method. I really want to know the calculus method. thank you

 

[HallsofIvy]

I am trying to find the area of the semi circle using integration given this velocity [ f'(x) ] graph

View attachment 4906

I figured out the equation of the circle but I don't know how to integrate it. Would trig substitution work for this problem since it uses coord points or am I just approaching it incorrectly?
I know how to do the geometry method. I really want to know the calculus method. thank you
View attachment 4907

You say you want to find the area of the semi-circle. With that "+ 2", you are finding the area below the semi-circle and the above the x-axis. That is, that will give you the area of the semi-circle plus the area of the rectangle it is "sitting on". I presume you know that the area of a circle of radius 2 is $\displaystyle 4\pi$ so the area of the semi-circle is $\displaystyle 2\pi$. The area of the "2 by 4" rectangle below it is 2(4)= 8. Use that to check your integration.

Yes, to do that integral, you should use a trig substitution. If you let $\displaystyle x= 2 sin(t)+ 10$, then $\displaystyle x- 10= 2 sin(t)$ so that $\displaystyle 4- (x- 10)^2= 4- 4sin^2(t)= 4(1- sin^2(t))= 4cos^2(t)$.
Of course, with $\displaystyle x- 10= 2 sin(t)$, $\displaystyle dx= 2 cos(t) dt$. When x= 8, 8- 10= -2= 2 sin(t) so sin(t)= -1, $\displaystyle t= -\pi$. When x= 12, 12- 10= 2= 2 sin(t), $\displaystyle t= \pi$. The integral becomes $\displaystyle 8 \int_{-\pi}^{\pi} cos^3(t) dt$. That can be done by writing the integrand as $\displaystyle cos^2(t)(cos(t)dt)= (1- sin^2(t)(cos(t)dt)$ and making the second substitution $\displaystyle u= sin(t)$.

The integral $\displaystyle \int_8^12 2dx= \left[2x\right]_8^{12}= 2(12)- 2(8)= 24- 16= 8$, the area of the 2 by 4 rectangle the semi-circle is sitting on.

 

[HallsofIvy]

You say you want to find the area of the semi-circle. With that "+ 2", you are finding the area below the semi-circle and the above the x-axis. That is, that will give you the area of the semi-circle plus the area of the rectangle it is "sitting on". I presume you know that the area of a circle of radius 2 is $\displaystyle 4\pi$ so the area of the semi-circle is $\displaystyle 2\pi$. The area of the "2 by 4" rectangle below it is 2(4)= 8. Use that to check your integration.

Yes, to do that integral, you should use a trig substitution. If you let $\displaystyle x= 2 sin(t)+ 10$, then $\displaystyle x- 10= 2 sin(t)$ so that $\displaystyle 4- (x- 10)^2= 4- 4sin^2(t)= 4(1- sin^2(t))= 4cos^2(t)$.
Of course, with $\displaystyle x- 10= 2 sin(t)$, $\displaystyle dx= 2 cos(t) dt$. When x= 8, 8- 10= -2= 2 sin(t) so sin(t)= -1, $\displaystyle t= -\pi$. When x= 12, 12- 10= 2= 2 sin(t), $\displaystyle t= \pi$. The integral becomes $\displaystyle 8 \int_{-\pi}^{\pi} cos^3(t) dt$.

Ishuda has pointed out a typo- I meant $\displaystyle cos^2(t)$.

That can be done by writing the integrand as $\displaystyle cos^2(t)(cos(t)dt)= (1- sin^2(t)(cos(t)dt)$ and making the second substitution $\displaystyle u= sin(t)$.

The integral $\displaystyle \int_8^12 2dx= \left[2x\right]_8^{12}= 2(12)- 2(8)= 24- 16= 8$, the area of the 2 by 4 rectangle the semi-circle is sitting on.