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 y is equals to x square, and the secon

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

Key Concepts

Setting Up Definite Integrals

A key step in solving area problems is correctly setting up the definite integrals. This involves identifying the correct bounds determined by the intersection of the curves and ensuring that the integrand represents the difference between the functions that bound the region. Mastery of this concept is essential for applying integration techniques to solve problems in calculus.

Integration with Respect to x and y

This concept explores the flexibility of integration in calculating areas. When integrating with respect to x, one considers vertical slices of the region, while integration with respect to y involves horizontal slices. Choosing the appropriate variable can simplify the computation by aligning the limits of integration with the geometry of the region.

Area between Curves

This concept involves finding the region enclosed by two functions. Typical methods subtract the lower function from the upper function when the curves are expressed in the same variable (usually x), or subtract the left function from the right function when expressed in y. Understanding how to compute the area between curves is central to applications of definite integrals in calculus.

Intersection Points

Determining the intersection points of the curves is crucial because these points define the limits of integration. By setting the equations equal to each other, one can find where the curves meet, and these values are used to properly bound the definite integrals whether integrating with respect to x or y.

Transcript

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

00:03 The first one is y is equals to x square, and the second one is y is equals to 6 minus x.

00:13 We are required to find the area of the region bounded by the graphs of these equations, with the help of integration, firstly with respect to x and after that with respect to y.

00:25 So let's see how to solve this question.

00:28 First of all, let's draw the graph for these functions, and the graph is shown below.

00:35 So this is the graph for these two equations.

00:38 This blue curve represents y is equal to x square and this green line represents y is equals to 6 minus x and this is the common region whose area is required to be found.

00:57 Now let's find the point of intersection by equating the above two equations so we can write x squared is equal to 6 minus x or we can write it as x square plus x minus 6 is equals to 0 this is a quadratic equation and now let's factorize this so we will have x plus 3 into x minus 2 is equal to 0 so x will be equals to minus 3 comma 2 hence we can say that the point of intersection are minus 3 .9 and 2 .4.

01:59 And now let's write the expression to calculate the area of the region.

02:06 So we will have area a is equals to minus 3 to 2, 6 minus x minus x squared dx.

02:19 And let's simplify this.

02:22 So we will have integration minus 3 to 2, 6 minus x minus x squared d x.

02:32 So now we have simplified this expression.

02:37 So let's integrate these terms now.

02:40 So we will have 6x minus the integration of x d x will be equal to x square upon 2 minus the integration of x squared d x will be equals to x cube upon 3...

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