To find the volume of a right circular cone with base radius r and height h, the cone is divided into n frustums of equal heights. The volume of each frustum is approximated as if it were a circular...

To find the volume of a right circular cone with base radius r and height h, the cone is divided into n frustums of equal heights. The volume of each frustum is approximated as if it were a circular...

Zero to infinity∑ (-1)n * ((5n2+7)/(7n2+10))^n I think I have to use the root test but not sure. What test do I use? If it is the root test, does the square root cancel...

Find the area bounded by the curve: y = x(x-1)^2, the line y = 2, and the y - axis

How do I do this ?

I dont understand this because when I find the intersection points which is x=0 and x=1 , how do I know which is the outer and lower function ?

Basically what the questions states. Thanks

I'm having a lot of trouble with the set up and integration. Please help me with this problem, thank you. I know you are supposed to use the formula integral(2pi * y * sqrt(1+(y')^2)).

∫ (1)/(sqrt(52x -1))dx Please show me steps. Im having a hard time figuring this out

I still havent learned anthing about logarithm derivatives yet so i have no idea how to proof this even i know the answer is ln2. So far with the definition of integral i transformed...

∫ba (2+x-x2)dx

2∫ab f(x)f'(x)dx=[f(b)]2-[f(a)]2

∫a b (2+x-x2)dx is a maximum. Please explain your reasoning.

lim 1 1 2 3 n — [ (—)9+(—)9+(—)9+...

∫01 f(x)dx = ∫01 f(1-x)dx

∫02 [x]nf'(x) dx Please hep me with this integration. It includes an integral part and a derivative :O

1.) Sketch the region enclosed by the curves below. 2.) Decide whether to integrate with respect to x or y. 3.) Find the area of the region. 2y=5sqrt(x), y=3, and 2y+1x...

I'm not sure how to start this one. All I know is sec^2 is (1/cosx)^2 is this right?

Let F(x) = ∫eu^2 + cos(u) du with upper limit = 7x, lowerlimit = 0 Compute F'(0) I have no idea how to approach this question the 7x in the upper limit is quite...

Evaluate the integral (1/sqrt(x^2 - 5))dx on limits of 3 to 4, using the formula: integral of (1/sqrt(x^2 - a^2))dx. Recall that this formula was found by using the substitution x = a cosh t,...

∫dx/(50-10x)=-1/10ln[50-10x]+c Why does factoring 1/10 a constant out of integral not work. ∫dx/(50-10x)=1/10∫dx/(5-x) u=5-x du=-1 -1/10ln[5-x]...

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