Video: APCALC02AB-P1A-Q28-805131059291

Calculate ∫_(1) ^(𝑒) ((ln π‘₯)⁴/π‘₯) dπ‘₯.

02:06

Video Transcript

Calculate the integral evaluated between one and 𝑒 of the natural log of π‘₯ to the fourth power divided by π‘₯ with respect to π‘₯.

We’re going to begin by rewriting our integral slightly. We can say that the natural log of π‘₯ to the fourth power divided by π‘₯ is the same as the natural log of π‘₯ of the fourth power times one over π‘₯. And then, we notice that the derivative of one part of our integral is equal to another part. The derivative of the natural log of π‘₯ with respect to π‘₯ is equal to one over π‘₯. This tells us we can use integration by substitution to solve this problem. We’re going to let 𝑒 be equal to the natural log of π‘₯. And we’ve then seen that the derivative of 𝑒 with respect to π‘₯ is equal to one over π‘₯.

Now, d𝑒 by dπ‘₯ is not a fraction. But for the purposes of integration by substitution, we can say that d𝑒 by dπ‘₯ equals one over π‘₯ is equivalent to d𝑒 equals one over π‘₯ dπ‘₯. And we can now see we can replace one over π‘₯ dπ‘₯ with d𝑒. And we can replace the natural log of π‘₯ with 𝑒. And we’ll be integrating 𝑒 to the fourth power with respect to 𝑒. But what about our limits?

Our limits are currently in terms of π‘₯. So we’ll use the fact that we let 𝑒 be equal to the natural log of π‘₯. And when π‘₯ is equal to 𝑒, we know that 𝑒 is going to be equal to the natural log of 𝑒, which is of course one. This also means that when π‘₯ is equal to one, 𝑒 is equal to the natural log of one which is of course zero. And so, we can now evaluate the integral between zero and one of 𝑒 to the fourth power with respect to 𝑒. To integrate 𝑒 to the fourth power, we add one to the power and divide by that new value.

So the integral of 𝑒 to the fourth power is 𝑒 to the fifth power divided by five. And we’re going to evaluate this between the limits of zero and one. That’s one to the fifth power divided by five minus zero to the fifth power divided by five which is equal to one-fifth. And we’re done.

We used 𝑒 is equal to the natural log of π‘₯ to change our limits. So we can see that the integral evaluated between one and 𝑒 of the natural log of π‘₯ to the fourth power divided by π‘₯ with respect to π‘₯ is equal to one-fifth.

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