Determine whether the following statements are true and give a...

Question

Determine whether the following statements are true and give an explanation or counterexample.

b. The area of the region between y=sin x and y=cos x on the interval [0,π/2] is ∫π/20(cosx−sinx)dx.

Accepted Answer

Welcome back, everyone, to another video. Consider the area between Y equals X cubed and Y equals 2 X on the interval from 0 to 2 inclusive. Is it correct to write the area as the integral from 0 to 2 of 2 X minus X cub to DX? A says yes because 2 x is always above X cubed. No, the integral should be split. C yes because X cubed is greater than 2 X on the interval from 0 to 2 inclusive and D no, because the curves do not intersect on that interval. So for this problem, let's recall that if we are constraining the area between two curves. We can calculate it as the integral from A to B, where A and B represent our interval. Of F of X minus G of XD X. This form is valid as long as F of X. Is the upper curve throughout the whole interval and G of X remains the lower curve, right? If they swap at some point, well, essentially we have to split our integral. So we can begin by finding the intersection points, if any, by setting X cubed equals to x. We can solve this equation by subtracting 2 X from both sides and factoring it out, so we get X multiplied by X2 minus 2 equals 0. This equation has two solutions. First of all, either X is equal to 0, right, because we have a product, or X2 minus 2 is equal to 0. So now we can see that there are actually 3 solutions, not 2, because we have a difference of X2 minus 2 equals 0, which can be written as the difference of squares in the form of x minus square root of 2 multiplied by X plus square root of 2 equals 0, right? So now we have 3 solutions for for the intersection points. X1 is 0, X2 is negative square root of 2, and X3 is positive square root of 2. On the given interval, we are only interested in 2 out of 3 solutions, those will be 0 and square root of 2 because they belong to the given interval. And essentially what we want to do is. Understand which curve is the upper curve, right? So, what we can do is simply take any point between 0 and square root of 2, and evaluate the value of each function. So for X cubed, we can take 1, right? Because 1 is between 0 and square root of 2. So that'd be Why? Equals 1 cubed, which is 1, and for 2 X. That'd be Y equals 2 multiplied by 1, which is 2. So which one is higher? Well, 2 X is equal to 2 at X equals 1. And 2 is greater than 1, which essentially means that 2 X is above X cubed from 0 to square root of 2. Now, what we're going to do is consider the interval from square root of 2 up to 2, right? Because now, square root of 2 is basically splitting our interval into two parts. We have 0 up to square root of 2. And then we have an interval from square root of 2. Up to 2, right? So, between square root of 2 and 2, we can take any points, such as, let's say square root of 3, or maybe we can take 1.5, right? Because square root of 2 is 1.4 approximately, so we can take 3 halves, and we can evaluate X cubed, which would be Y equals 3 halves cubed. And this is 27 divided by 8. And for 2 acts we can sake. The same 0.3 halves, that'd be 2 multiplied by 3 halves, and this is equal to 6 divided by 2, which is 3. So, which one is higher? 27 divided by 8 gives us a value over 3. So now X cubed is greater than 2 X, meaning our curves swap, right? So, the actual area can be written as the integral from 0 up to square root of 2. Of the difference between 2 X and X cubed DX because 2 X is the upward curve, we have a higher value. Plus the integral from square root of 2 up to 2 of X cubed minus 2 XD X, right? So we have to split our integral into two parts because we have an intersection points. We have an intersection point, I'm sorry, at which our curves swap, right? So the upper curve becomes the lower curve after square root of 2. Therefore, we can conclude that the correct answer is B. No, the integral should be split. Thank you for watching.

Determine whether the following statements are true and give an explanation or counterexample.b. The area of the region between y=sin x and y=cos x on the interval [0,π/2] is ∫π/20(cosx−sinx)dx.

Watch next

Determine whether the following statements are true and give an explanation or counterexample.

b. The area of the region between y=sin x and y=cos x on the interval [0,π/2] is ∫π/20(cosx−sinx)dx.