# Video: MATH-DIFF-INT-2018-S1-Q15B

Find β«(π₯ β 3)π^2π₯ dπ₯.

05:03

### Video Transcript

Find the integral of π₯ minus three timesed by π to the power of two π₯ with respect to π₯.

We can solve this integration by using integration by parts. Now, we know this because we have a polynomial which is π₯ minus three multiplied by an exponential which is π to the two π₯. And when performing integration by parts, we integrate one part of our integral and differentiate the other in order to form a new integral. And so, when we differentiate the polynomialβso thatβs π₯ minus threeβit will reduce. However, when we integrate the exponentialβthatβs π to the two π₯βit will only change by a constant and so it will not get any more difficult to integrate. Now, letβs quickly remind ourselves how integration by parts works.

If we have an integral of π’ multiplied by dπ£ by dπ₯ with respect to π₯, where π’ and dπ£ can be anything that we choose, then this integral is equal to π’π£ minus the integral of π£ timesed by dπ’ by dπ₯ with respect to π₯. So as we can see, the π’ from the original integral has been differentiated into the second integral to dπ’ by dπ₯. And the dπ£ by dπ₯ from the original integral has been integrated to π£ in the second integral. Now, letβs perform integration by part on our integral.

Weβll recall π₯ minus three π’. And π to the two π₯ will be dπ£ by dπ₯. Now, we need to find dπ’ by dπ₯ and π£. dπ’ by dπ₯ is the differential of π’ with respect to π₯. So we differentiate π₯ minus three with respect to π₯. In order to differentiate the π₯, we multiply by the power which is one since π₯ is equal to π₯ to the power of one and then reduce the power by one until we get π₯ to the zero. And π₯ to the zero is simply equal to one, giving us one multiplied by one. Then, since negative three is a constant and weβre differentiating it, this goes to zero. Since when differentiating any constant, we end up with zero. So therefore, dπ’ by dπ₯ is equal to one times one plus zero. So dπ’ by dπ₯ is simply equal to one.

Next, we need to find π£. And in order to do this, we simply integrate dπ£ by dπ₯, which is also equal to π to the power of two π₯. Weβll need to use the reverse chain rule in order to integrate this. We have that the integral of π to the power of π₯ with respect to π₯ is equal to π to the power of π₯. And so, when weβre integrating π to the power of two π₯, weβll get one divided by the differential of two π₯ which is two multiplied by π to the power of two π₯. This gives us one-half π to the power of two π₯.

Now remember that this two from the half has come from using the inverse chain rule. We have divided by the differential of two π₯. Now, we have found our π’, dπ’ by dπ₯, dπ£ by dπ₯, and π£. Weβre ready to use integration by parts. We have that the integral of π₯ minus three timesed by π to the two π₯ with respect to π₯ is equal to π’ timesed by π£. So thatβs π₯ minus three times one-half π to the two π₯ minus the integral of π£ dπ’ by dπ₯ with respect to π₯. So thatβs minus the integral of one-half π to the two π₯ times one with respect to π₯.

Now, this can be simplified to give π₯ minus three over two times π to the power of two π₯. And then, in the integral, we can ignore the times one since this has no effect in. And since one-half is a constant, we can bring it to the front of the integral giving us minus one-half times the integral of π to the two π₯ with respect to π₯.

Now, weβve already found the integral of π to the two π₯ with respect to π₯ since this is equal to π£. And we found that itβs equal to one-half π to the two π₯. This gives us π₯ minus three over two π to the two π₯ minus one-half times one-half π to the two π₯. And now, since this is the last integration we will perform, we mustnβt forget to add on our constant of integration or π, giving us plus π on the end. Next, we have one-half times one-half π to the two π₯. So we can write this as one-quarter π to the two π₯. And then, we notice that we have some π to two π₯ terms which we can group together.

In order to do this, we need to separate the fraction π₯ minus three over two into π₯ over two minus three over two. This gives us π₯ over two π the two π₯ minus three over two π to the two π₯ minus one-quarter π to the two π₯ plus π. When we simplify this, we find our solution which is that the integral of π₯ minus three π to the two π₯ with respect to π₯ is equal to π₯ over two times π to the two π₯ minus seven over four times π to the two π₯ plus π.