Name: Math 1520 In-Class Activity 2 1. Set up an integral...

Name: Math 1520 In-Class Activity 2 1. Set up an integral for the area of the shaded region. Evaluate the integral to find the area of the shaded region. y y = 5x - x$^2$ 6 4 (4, 4) y = 5x - x$^2$ = x(5 - x) = 0 x = 0 x = 5 A = \int_0^5 (5x - x$^2$)dx A = [\frac{5}{2}x$^2$ - \frac{1}{3}x$^3$]$^5_0$ = \frac{5}{2}(5$^2$) - \frac{1}{3}(5$^3$) = \frac{5 \times 25}{2} - \frac{1}{3}(125) = \frac{125}{2} - \frac{125}{3} = \frac{375 - 250}{6} y = x x -1 1 2 3 4 5 6

Transcript

00:02 To find the total area of the shaded region, which is really the area between y equals the square of x minus 3 and the x -axis, we're going to need to set up two different integrals.

00:17 The reason we're going to have to set up two different integrals is because this first region, which i'm going to outline or highlight here in red, this definite integral is going to give us a negative value.

00:30 So we're going to need to make that work out to be positive if we're going to find an area.

00:36 The other thing we're going to notice is that there is a point of intersection with the x -axis that we need to make a determination of to find out where our two integrals are going to separate.

00:49 So we're going to take the square root of x minus 3.

00:51 We're going to find out where that equals 0.

00:55 And what that's going to give us is x is equal to 9.

00:59 So this is the point 9 comma 0, so x value is 9.

01:04 So what that means is we're going to have our two integrals separated at 9 for our sort of our midpoint.

01:12 So if we find the area of this first area, we know that we need to take the opposite of the integral from 0 to 9 of the square root of x minus 3.

01:26 The reason we do that once again is that if i just take the definite integral from 0 ,000, 0 to 9, that's going to be a negative value.

01:34 By making it the opposite of that, that's going to allow us to get a positive number.

01:39 To that, we're going to add the integral from 9 to 16 of the square of x minus 3.

01:46 We don't have to worry about anything because here this area is above the x -axis, so it is already going to be positive.

01:54 If we treat the square root of x as x to the 1⁄2 power, that makes our...

02:02 Integration process a little bit easier.

02:06 So we're going to go ahead and make that change right here.

02:13 And then we'll go ahead and find our integral.

02:17 In this case, what we see is that we've got x to the three halves divided by three halves or two -thirds x to the three -haves minus three -x.

02:28 That's going to be evaluated from zero to nine.

02:30 And then since it's the same integrand, we're going to have the same integral.

02:39 The difference here, of course, is we're evaluating from 9 to 16...

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