Finding area of region bounded (Integration)

Hi, so I have this problem:

To find the area of the region bounded by graphs of y=x^{3}, y=0, x=1, and x=3 by limit definition

But I was a bit confused on this because I didn't really know what the intervals were.

Like I graphed it out, and it would look like this:

I will describe it: A horizontal line on y=0, then 2 veritcal lines on x=1 and x=3, then the x^{3} curves from down to up, hits the 0, but the shape looks pretty weird.

Can someone explain to me what area/intervals are supposed to be?

From my intepretation when I saw the graph of all those equations, it looked like it would be [1, 3] since 3 looks like the max, and 1 is like the min, and there's where they all come together

They make somewhat like a rectangle at the bottom, then it curves up, so like a curvy trapezoid? Maybe?

Re: Finding area of region bounded (Integration)

I think you have the region right - its boundaries are the vertical lines $\displaystyle x=1$ and $\displaystyle x=3$ on the right and left, the horizontal line $\displaystyle y=0$ on the bottom, and the curve $\displaystyle y=x^3$ on the top.

The way I learned it was to divide the region into vertical rectangles. These rectangles have width $\displaystyle dx$ and height $\displaystyle x^3$, so the sum of $\displaystyle x^3\,dx$ is the area of the region. In the limit, it's the integral of $\displaystyle x^3\,dx$ and the limits are $\displaystyle x=1$ on the left and $\displaystyle x=3$ on the right. So the area is given by:

$\displaystyle \int_1^3x^3\,dx$.

Hopefully that answers your question.

- Hollywood

Re: Finding area of region bounded (Integration)

Quote:

Originally Posted by

**hollywood** I think you have the region right - its boundaries are the vertical lines $\displaystyle x=1$ and $\displaystyle x=3$ on the right and left, the horizontal line $\displaystyle y=0$ on the bottom, and the curve $\displaystyle y=x^3$ on the top.

The way I learned it was to divide the region into vertical rectangles. These rectangles have width $\displaystyle dx$ and height $\displaystyle x^3$, so the sum of $\displaystyle x^3\,dx$ is the area of the region. In the limit, it's the integral of $\displaystyle x^3\,dx$ and the limits are $\displaystyle x=1$ on the left and $\displaystyle x=3$ on the right. So the area is given by:

$\displaystyle \int_1^3x^3\,dx$.

Hopefully that answers your question.

- Hollywood

Ooo

I see.

So the two x's are the boundaries, and the slope is x^3

Not sure where the y=0 goes.

But I think I get it, thanks :)