Can someone help me with some Calculus AB problems?

Question

1) Create an integral that gives the exact value for (1^2+2^2+3^2+4^2+...+n^2)/(n^3)

2) Find the limit, as x approaches 5, of (2x/(x-5)) x the integral of [sin(t)]/t dt.
I think this has something to do with as f(a)-f(x)/(a-x)=f'(a) or something like that. I heard that it is 0 or undefined.

3) Evaluate the integral of square root of (b^2-x^2) dx from the interval (b/square root of 2) to b. I need to show the graph and find an exact value without using a calculator.

You don't have to answer all questions but 1 would be nice.

1) Create an integral that gives the exact value for
(1^2+2^2+3^2+4^2+...+n^2)/(n^3).

2) Find the limit, as x approaches 5, of (2x/(x-5)) x the integral of [sin(t)]/t dt.
I think this has something to do with as f(a)-f(x)/(a-x)=f'(a) or something like that. I heard that it is 0 or undefined.

3) Evaluate the integral of square root of (b^2-x^2) dx from the interval (b/square root of 2) to b. I need to show the graph and find an exact value without using a calculator.

You don't have to answer all questions but 1 would be nice.

-->(1^2+2^2+3^2+4^2+...+n^2)/(n^3)

(1^2+2^2+3^2+4^2+...+n^2)/
(n^3)

Answer 1

1) Your question got cut off, please type it with spaces in between terms so the whole thing can be seen

2) You can't take the limit of an indefinite integral, because it has an arbitrary constant. Shouldn't that integral have limits?

3) Use the substitution x = b sin(t). Then dx = b cos(t) dt, and sqrt(b^2-x^2) = b cos(t) as well. Therefore, your new integrand is b^2 cos^2(t) dt. The new limits are pi/4 and pi/2, obtained by plugging in x=b/2 and x=b into x=b sin(t). To integrate, rewrite cos^2(t) = (1+cos(2t))/2.

Answer 2

Integrating the function between x = 0 and x = 2 is complicated as you're having to combine a logarithmic function. So, re-manage and combine in the different course. If y = 4ln(3 - x) then, y/4 = ln(3 - x) so, e^(y/4) = 3 - x => x = 3 - e^(y/4) Integrating w.r.t. y provides: 3y - 4e^(y/4) from y = 0 to y = 4ln3 Substituting provides: [12ln3 - 12] - [0 - 4] = 12ln3 - 8 Now, region R is the part of the rectangle minus the above for this reason: (6 x 2) - (12ln3 - 8) = 20 - 12ln3 = 6.817 Have a flow at something and enable me comprehend in case you warfare......i admire your philosophy on learn strategies!! :)>