35-38. Area and volume Let R be the region in the first quadra...

Question

35-38. Area and volume Let R be the region in the first quadrant bounded by the graph of

Piecewise function f(x) equals 1 for x between 0 and 1, and 2 minus square root of x for x between 1 and 4.

Find the area of the region R.

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says find the area of the region R in the first quadrant embedded by the curve F of X, which is equal to 2, if 0 is less than or equal to X, which is less than or equal to 1, and 3 minus the square root of X, if 1 is less than X, which is less than or equal to 9. Below the problem we're given a graph that has our function F of X plotted on it. It also has a region R shaded on this graph, and R, the region is bounded above by our function F of X, bounded below by the X-axis and embedded on the left by the Y axis. We're also given 4 possible choices as our answers. For choice A, we have 26 divided by 7 square units. For choice B, we have 29 divided by 3 square units. For choice C, we have 26 divided by 3 square units, and for choice D, we have 26 divided by 5 square units. So we're asked to find the area of the region R in the 1st 1st quadrant. And if we look at our graph, we see that the area of region R corresponds to the area underneath the curve of F of X. So, to find this area, we just need to integrate F of X. Now, in order to find the upper limit of our limits of integration, we need to find the point where F of X intersects the X axis. And looking at a graph, this appears to occur when X is equal to 9, but let's see this analytically. So in order to find the X intercept, we need to set F of X equal to 0, and so we're going to have 0 is equal to 3 minus the square root of X. So here we've used the second piece of our piece-wise function since our first piece is a constant that's equal to 2 and can never be zero. And so now we need to solve this equation for X. So the first step is to add the square root of X to both sides. So we'll get the square root of X is equal to 3, and square on both sides we'll see that X is going to be equal to 9. So that means that our area A is going to be equal to the interval from 0 to 9 of F X DX. Now, since F of X is a piecewise function, we'll need to break this up into two different integrals. So that means our area is going to be equal to. For the first piece we'll have the integral from 0 to 1 of 2 DX. Then we'll have plus the integral from 1 to 9 of our second piece, which is going to be the quantity of 3 minus X-rays to the 1/2. In quantity, DX. So here we just rewritten the squared X as X to 12 power because it'll be easier to integrate in this form. So, perform the integration, we see that our area is going to be equal to. For the first interval, when we integrate 2 with respect to X, we get 2 X, and that's going to be evaluated from 0 to 1. Then for the second interval, we'll have plus the quantity, and for the second interval, we'll integrate term by term. So the first term, we're going to be integrating 3 with respect to X, which will give us 3 X. Then we'll have minus, and for the second term, we're integrating X to the 1/2 power with respect to X, and that will give us X to the 3/2 power divided by the quantity of 3 divided by 2 in quantity. And this quantity needs to be evaluated from 1 to 9. So, substituting in our limits of integration, we say our area is going to be equal to 2, multiplied by the quantity of 1 minus 0 in quantity. Plus 3 multiplied by the quantity of 9 minus 1 in quantity minus 2 divided by 3 multiplied by the quantity of 9 raised to the 3 halves power minus 1 raised to the 3 halves power. In quantity. So simplifying this expression, this is going to be equal to. For the first two terms, we're going to have 2 plus 24, which is going to give us 26. Then we'll have minus 2 divided by 3, multiplied by, and we have 9 rates to the 3 halves. Power minus 1 to the 3/2 power, that's going to also give us 26. So we can factor out 26 out of both of those terms. So that means our area is going to be equal to 26, multiplied by the quantity of 1 minus 2 divided by 3 in quantity. And when we simplify this expression, we see that our area is going to be equal to 26 divided by 3 square units. So that is the answer that we are looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful and I'll see you in the next video.

35-38. Area and volume Let R be the region in the first quadrant bounded by the graph of

Find the area of the region R.

Key Concepts

Piecewise Functions

Definite Integrals for Area Calculation

Square Root Functions and Their Integration

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