Find the area of the region by integrating (a) with respect to x and (b) with respect to y.

step 1 In this question, we have two equations. The first one is x is equals to 4 minus y square, and t

Find the area of the region by integrating (a) with respect...

Key Concepts

Area Between Curves

This concept involves finding the area of a region that is bounded by two curves. The area is determined by setting up an integral where one curve represents the upper or right boundary and the other represents the lower or left boundary, and then integrating the difference between these functions over the appropriate limits.

Integration With Respect to a Variable

This refers to setting up and evaluating integrals in either the x-direction or y-direction. Integrating with respect to x (dx) or y (dy) changes how the region's boundaries are expressed and often requires solving the given functions for the other variable, which affects the limits of integration.

Determining Limits of Integration

This involves finding the points of intersection between the curves to establish the boundaries for the definite integral. The intersection points are essential as they define the exact interval over which the integration is performed, ensuring that the entire region of interest is covered.

Changing the Order of Integration

This key idea is about setting up the area integral by integrating with respect to either x or y. When you change the integration order, you must reinterpret the curves as functions of the new variable and adjust the integration limits accordingly. This technique allows for flexibility depending on the form of the given functions.

Transcript

00:01 In this question, we have two equations.

00:03 The first one is x is equals to 4 minus y square, and the second one is x is equals to y minus 2.

00:12 We are required to find the area of the region bounded by the graphs of these equations by integration with respect to x and with respect to y.

00:22 So let's see how to solve this question.

00:26 First of all, let's draw the graph for these equation and the graph is shown below.

00:30 So this is the graph for these equations.

00:34 This red curve represents x is equal to 4 minus y square and this blue line represents x is equals to y minus 2 and the common region whose area is required to be found is this one.

00:56 From the first equation now we can write y is equals to plus minus under root 4 minus x and from the second equation we can write y is equal to x plus 2 and now let's find the point of intersection and to do so equate both of the equations so we can write 4 minus y square is equal to y minus 2 or this will be equals to y square plus y minus 6 is equal to 0 it is a quadratic equation so now let's factor this we can write y plus 3 into y minus 2 is equal to 0 so from here we will have y is equals to minus 3 comma 2 so from here we can conclude that the point of intersection are minus 5 comma minus 3 and 0 .2 and now on on the basis of this information, now let's write the expression to calculate the area of the region.

02:36 So, a will be equals to integration minus 5 to 0 x plus 2 minus minus under root 4 minus x d x plus integration 0 to 4 2 into under root 4 minus x d x.

03:04 Now let's simplify this.

03:06 So we can write integration minus 5 to 0, x plus 2 plus under root of 4 minus x d x plus 2 plus 2 into integration 0 to 4 under root of 4 minus x d x.

03:32 Now let's integrate these values.

03:35 So we will have the integration of x dx will be equal to x squared upon 2 plus the integration of 2 d x will be equal to 2x minus 2 by 3, 4 minus x to the power 3 by 2 and the limits are minus 5 to 0 plus 2.

04:02 And the integration of under root 4 minus x t x will be equals to minus 2 upon 3, 4 minus x to the power 3 upon 2.

04:14 And the limits are 0 to 4.

04:17 Now, substitute the limits.

04:22 So, we get 0 plus 0 minus 2 by 3 into 4 to the power 3 by 2 minus 25 upon 2...

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