00:01
In this question, we are asked to find the area of the region under the graph of f and over the given interval.
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First, let's sketch the graph of the region.
00:17
To sketch the graph of x to the 1 third, we basically need to draw the graph of x -cube and by points when x equals 0, when x equals 1 is 1, when x equals 2, x -cube equals to 8.
00:44
So basically this is a graph of y equals x cube.
00:57
And to get the graph, so this is y equals x cube.
01:01
And to get the graph of x to the one third, we simply need to rotate y equals x cube around the origin, around the line y equals to x.
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And if you rotate it around the line y equals to x, what you are going to get is this graph here.
01:34
So this graph here is the graph of y equals to x to the one third.
01:39
And we're interested in the part from x equals to 1 to x equals to 8.
01:51
So we are interested, we are asked to find this area here.
01:58
And to do that, we need to calculate the integral from 1 to 8 of f of x d x.
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So we need to integrate x to the 1 3rd from 1 to 8.
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And by the power rule, the antiderivative of x to the 1 3rd equals to x to the 1 3.
02:23
1 3rd plus 1, divide by 1 3rd plus 1...