{Use of Tech} Using the integral of sec³u By reduction formula...

Question

{Use of Tech} Using the integral of sec³u By reduction formula 4 in Section 8.3,

∫sec³u du = 1/2 (sec u tan u + ln |sec u + tan u|) + C

Graph the following functions and find the area under the curve on the given interval.

f(x) = 1/(x√(x² - 36)), [12/√3 , 12]

Accepted Answer

Hello, in this video, we are going to be finding the area given by the shaded region. So we are given the shaded region where the upper curve is defined as F of X is equal to 25 divided by X multiplied by the square root of X2 minus 25. And this is a region that spans between 6 and 10 along the x-axis. So the easiest way that we can find this area is to integrate the area with respect to X. The area is going to be defined as the integral from 6 to 10 of the uppermost curve minus the lowermost curve, which is going to be given as X is equal to 0. So that is just going to be the integral from 6 to 10 of 25 divided by X multiplied by the square root of X2 minus 25, and we will be integrating with respect to X. Now we can go ahead and solve this integral by using trigonometric substitution. If we allow X to equal to 5 sequin theta. Then DX is going to be 5 seeking theta tangent of theta. The theater. And here, the value of theta is going to equal to sequent inverse of X divided by 5, which we will set to the side for now. Now, if we were to apply these trigonometric substitutions, we will have to change the bounds with respect to X. So plugging in the value of X into each of the bounds is going to give us an upper bound, which we will label as A or B. And our upper bound is going to be pi divided by 3. And plugging in 6, the lower bound is going to give us a lower bound of cosine inverse of 5 divided by 6. So the area is going to equal to the integral from A to B. Of 25 divided By 5 sequent of theta multiplied by the square root. Of 5 seekin of theta squared. -25 multiplied by the quantity, 5 sequence of theta. Tangent of theum de theta. Now what we need to do is we need to go ahead and simplify. So if we are taking a look at the given square root, if we were to square the given quantity, we will have the square root of 25 square theta minus 25. If we were to factor out a 25 and pull that in front of the square root, we will have 5 multiplied by the square root of seeking square theta minus 1. By using An identity for seek and square theta minus 1, we can rewrite this as the square root of tangent square theta. And this is going to further simplify to leave us with 5 tangent of data. So, we can go ahead and simplify the given integral to be the integral from A to B of 25, multiplied by 5, multiplied by seeking theta, multiplied by tangent of theta, and this is going to be divided by 5, multiplied by seeking of theta, multiplied by 5, tangent of theta D theta. Now what we need to do is simplify. We have a factor of 5 that we can eliminate from both the numerator and denominator. We have a factor of seeking that we can eliminate and a factor of tangent, and 5 divided by 25 is going to simplify to be 5 divided by 1. So this integral will further simplify to be the integral from A to B of 5D theta. And the antiderivative of this is going to be 5 theta evaluated from A to B. Now, again, we said that our lower bound is going to be cosine inverse of 5 divided by 6, and our upper bound is pi divided by 3. So 5 theta will be evaluated from cosine inverse of 5 divided by 6 to pi divided by 3. And by expanding this, we will have 5 multiplied by the quantity, pi divided by 3 minus cosine inverse of 5 divided by 6, and this is going to be the solution. And that is going to be the area. That is given to us in the graph. And the overall solution is going to be B. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

{Use of Tech} Using the integral of sec³u By reduction formula 4 in Section 8.3,
∫sec³u du = 1/2 (sec u tan u + ln |sec u + tan u|) + C

Graph the following functions and find the area under the curve on the given interval.
f(x) = 1/(x√(x² - 36)), [12/√3 , 12]

Key Concepts

Integration Using Reduction Formulas

Definite Integrals and Area Under a Curve

Trigonometric Substitution in Integration

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