(a) Approximate the area (in square units) under the curve...

(a) Approximate the area (in square units) under the curve $f(x) = \frac{1}{2}x^2$ from 0 to 2 using 10 left rectangles with equal bases. (Round your answer to three decimal places.) (b) Find the exact area (in square units) under the curve between the given x-values by evaluating an appropriate definite integral using the Fundamental Theorem.

Transcript

00:01 All right, let's say we have a function f of x equals 4 over x.

00:06 And we're interested in the area under this curve between values a equals 1 and b equals 3.

00:14 And we want to start by using a riemann sum of 10 rectangles.

00:18 So first, let's draw this just to get an idea of what we're dealing with here.

00:23 So let's just draw x, y axis here.

00:25 We know that we're going from 1 to 3.

00:30 And our function, even if we don't have any idea of what the function actually looks like, it's fine.

00:35 But i'll just draw a function that sort of looks like this.

00:39 All right.

00:40 And if we want 10 rectangles, that means that we need to divide this space between 1 and 3 into 10 rectangles, all with the width of delta x.

00:52 And the formula for delta x, which is the width of each rectangle, is b minus a, which is the distance from 3 to 1.

01:03 To mark off, this is a equals 1, b equals 3, over n, which is the number of rectangles.

01:10 And so that means we have 3 minus 1 over 10.

01:16 And this is going to equal to 2 over 10, which is equal to 0 .2.

01:24 So that means each one of our rectangles, we're going to start here.

01:28 And each one of them is going to be delta x equals 0 .2 long.

01:36 So i'm actually going to draw my rectangle.

01:38 There's a couple different ways to do a riemann sum.

01:40 I'm going to do what's called a left -handed sum, meaning i'm going to start with my line connecting on the left side.

01:48 So my function meets with the left side of the rectangle.

01:55 So this value here that's in the dot is actually f of 1.

02:00 All right.

02:00 This is the function evaluated at 1.

02:03 My next rectangle here, again, is going to have that same width of 0 .2.

02:12 Actually, i'll just write these as delta x, since we already know that delta x is 0 .2.

02:19 So this is going to be delta x here, another delta x here.

02:24 And this value here now is going to be f of 1 .2.

02:30 Because remember, we've gone 0 .2 over to the right.

02:34 So now, instead of our dot being at f of 1, it'll be f of 1 .2...

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