Exact value of the are of the region bounded by

[kylebutler]

Question:
Find the exact value of the are of the region bounded by:
x^3, the x-axis and x=1 and x=4

Answer is 3.75

I tried finding the anti derivative so 1/4(x)^4, and therefore I've got 1/4(4)^4 - 1/4, which isn't the correct answer

 

[Dick]

You are integrating x^3 for x from 0 to 1, right? So the antiderivative is, yes, x^4/4 and the result is 1^4/4-0^4/4=1/4. That's what you are doing yes? You've got a bunch of what I assume are 'dashes' in the post that look like 'minuses'. That looks ok to me, and I don't see how 3.75 could be correct.

 

[kylebutler]

I apologize, I meant x=1 and x=4 and I deleted all the dash (there was just one)

 

[Dick]

I apologize, I meant x=1 and x=4

That's different! So you did evaluate the antiderivative between 4 and 1. Then 4^4/4-1/4 is correct. And it's not 3.75. It's 63.75. A typo?

 

[kylebutler]

the answer in the book clearly states "3 and 3/4 units^2". If the answer is wrong than that's pretty awful, since it was the first problem. Other answers seem fine. Thanks for your help though.

If you don't mind another question I have is
Find the exact value of the are of the region bounded by:
e^x, the x-axis, the y-axis and x=1

I have no what to do if its bounded by both x and y axes

 

[Dick]

the answer in the book clearly states "3 and 3/4 units^2". If the answer is wrong than that's pretty awful, since it was the first problem. Other answers seem fine. Thanks for your help though.

If you don't mind another question I have is
Find the exact value of the are of the region bounded by:
e^x, the x-axis, the y-axis and x=1

I have no what to do if its bounded by both x and y axes

The 'y-axis' is x=0, right? That makes it pretty similar to the previous question.