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To find the area of the cylinder we need to find its volume first. Remember that the formula for the volume of a cylinder is 
where:
is the volume
is the radius
is the height
From the question we know that
and 
. Lets replace those values in our volume formula:
Now we can solve for to find our radius:
Now that we know the radius, we can use the formula for the area of a cylinder
where:
is the area
is the radius
is the height
We know now that and , so lets replace those values in our area formula:
We can conclude that the area of a cylinder that has a volume of 175 cubic units and a height of 7 units is 174.07 square units.
where:
From the question we know that
Now we can solve for
Now that we know the radius, we can use the formula for the area of a cylinder
where:
We know now that
We can conclude that the area of a cylinder that has a volume of 175 cubic units and a height of 7 units is 174.07 square units.
Hello,
a)
![I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\ = [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\ ](https://tex.z-dn.net/?f=I%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%2Asin%5E%7Bn-1%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%5B-cos%28x%29%2Asin%5E%7Bn-1%7D%28x%29%5D_0%5E%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2B%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%28x%29%2Asin%5E%7Bn-2%7D%28x%29%2Acos%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D0%20%2B%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%5E2%28x%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7B%281-sin%5E2%28x%29%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20-%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%20%5C%2C%20dx%5C%5C%0A%0A)
b)
![\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\ = \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\ = \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\ ](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7B3%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cdfrac%7B2%7D%7B3%7D%5C%20%5B-cos%28x%29%5D_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%3D%5Cdfrac%7B2%7D%7B3%7D%20%5C%5C%0A%0A%0A%0A%0A)
c)
a)
b)
c)
The average rate of change of the function at the interval is -8
<h3>How to determine the average rate of change?</h3>The table that completes the question is added as an attachment
From the table, we have:
f(5) = -26
f(3) = -10
The average rate is then calculated as:
Rate = (f(5) - f(3))/(5 -3)
This gives
Rate = (-26 + 10)/(5 -3)
Evaluate
Rate = -8
Hence, the average rate of change of the function is -8
Read more about average rate of change at:
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