# Thread: Finding area of region bounded (Integration)

1. ## Finding area of region bounded (Integration)

Hi, so I have this problem:
To find the area of the region bounded by graphs of y=x3, 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 x3 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?

2. ## Re: Finding area of region bounded (Integration)

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

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

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

- Hollywood

3. ## Re: Finding area of region bounded (Integration)

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

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

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