Finding area between two curves

[cem348]

I think that I am solving this problem correctly, but I can't get my answer to match the correct response. Thanks for any help!

Find the area of the region bounded by the curves y=f(x)= x+3 and y=g(x)= 4 times the square root of x.
(Answer=5.333)

I first graphed them on the calculator to see that g(x) was the upper boundary. I set the f(x) and g(x) equal to one another and then solved to get intersection points of 1 and 9. I squared the f(x) and g(x) to get rid of the sq root. So then I subtracted f(x) from g(x) to get 10x - xsquared -9. I integrated this to get 5xsquared minus (1/3)xcubed minus 9x. In plugging in the 9 and then trying to subtract with the 1 plugged in, I keep getting 81 + 4.3 for an answer of 85.3 although the answer is 5.333

 

[Unco]

I think that I am solving this problem correctly, but I can't get my answer to match the correct response. Thanks for any help!

Find the area of the region bounded by the curves y=f(x)= x+3 and y=g(x)= 4 times the square root of x.
(Answer=5.333)

I first graphed them on the calculator to see that g(x) was the upper boundary. I set the f(x) and g(x) equal to one another and then solved to get intersection points of 1 and 9.
So far so good.

I squared the f(x) and g(x) to get rid of the sq root. So then I subtracted f(x) from g(x) to get 10x - xsquared -9.
Not so good. Clearly f(x) - g(x) is not equal to (f(x))^2 - (g(x))^2.

Instead, you have to get used to square roots. \(\displaystyle \mbox{\sqrt{x} = x^{\frac{1}{2}}}\), which is no more difficult to integrate than \(\displaystyle \mbox{x^2}\).

I integrated this to get 5xsquared minus (1/3)xcubed minus 9x. In plugging in the 9 and then trying to subtract with the 1 plugged in, I keep getting 81 + 4.3 for an answer of 85.3 although the answer is 5.333

 

[soroban]

Hello, cem348!

Find the area of the region bounded by the curves: \(\displaystyle f(x)\:=\:y \:=\: x\,+\,3\) and \(\displaystyle g(x)\:=\:y\:=\:4\sqrt{x}\)

Your "squaring" is incorrect.

The area lies between \(\displaystyle x\,=\,1\) and \(\displaystyle x\,=\,9\) with \(\displaystyle g(x)\) on top.

The integral is: \(\displaystyle \L\:A \;=\;\int^{\;\;\;9}_1\left[4x^{\frac{1}{2}} - (x + 3)\right]\,dx\)