Area bounded by y2=x-1 and the line y=x-3

Area bounded by y2=x-1 and the line y=x-3

The question asks to estimate the area under the graph given the following table: | x | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 | | y | 4 | 10 | 8 | 6 | 14 | 10 | 12 | a) Use Simpson's...

A ball drops from a height of 20 feet. Each time it hits the ground, it bounces up 40 percents of the height it fall. Assume it goes on forever, find the total distance it travels.

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...

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 ?

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

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

Basically what the questions states. Thanks

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

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.

How do I do this ?

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)).

lim 1 1 2 3 n — [ (—)9+(—)9+(—)9+...

∫01 f(x)dx = ∫01 f(1-x)dx

∫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]...

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...

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...

∫ba (2+x-x2)dx

∫dx/(50-10x)=? Does factoring out 1/10 change the answer.