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

Find ∫(π‘₯ βˆ’ 3)𝑒^2π‘₯ dπ‘₯.

05:03

Video Transcript

Find the integral of π‘₯ minus three timesed by 𝑒 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 𝑒 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 that’s π‘₯ minus threeβ€”it will reduce. However, when we integrate the exponentialβ€”that’s 𝑒 to the two π‘₯β€”it will only change by a constant and so it will not get any more difficult to integrate. Now, let’s quickly remind ourselves how integration by parts works.

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

We’ll recall π‘₯ minus three 𝑒. And 𝑒 to the two π‘₯ will be d𝑣 by dπ‘₯. Now, we need to find d𝑒 by dπ‘₯ and 𝑣. d𝑒 by dπ‘₯ is the differential of 𝑒 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 we’re differentiating it, this goes to zero. Since when differentiating any constant, we end up with zero. So therefore, d𝑒 by dπ‘₯ is equal to one times one plus zero. So d𝑒 by dπ‘₯ is simply equal to one.

Next, we need to find 𝑣. And in order to do this, we simply integrate d𝑣 by dπ‘₯, which is also equal to 𝑒 to the power of two π‘₯. We’ll need to use the reverse chain rule in order to integrate this. We have that the integral of 𝑒 to the power of π‘₯ with respect to π‘₯ is equal to 𝑒 to the power of π‘₯. And so, when we’re integrating 𝑒 to the power of two π‘₯, we’ll get one divided by the differential of two π‘₯ which is two multiplied by 𝑒 to the power of two π‘₯. This gives us one-half 𝑒 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 𝑒, d𝑒 by dπ‘₯, d𝑣 by dπ‘₯, and 𝑣. We’re ready to use integration by parts. We have that the integral of π‘₯ minus three timesed by 𝑒 to the two π‘₯ with respect to π‘₯ is equal to 𝑒 timesed by 𝑣. So that’s π‘₯ minus three times one-half 𝑒 to the two π‘₯ minus the integral of 𝑣 d𝑒 by dπ‘₯ with respect to π‘₯. So that’s minus the integral of one-half 𝑒 to the two π‘₯ times one with respect to π‘₯.

Now, this can be simplified to give π‘₯ minus three over two times 𝑒 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 𝑒 to the two π‘₯ with respect to π‘₯.

Now, we’ve already found the integral of 𝑒 to the two π‘₯ with respect to π‘₯ since this is equal to 𝑣. And we found that it’s equal to one-half 𝑒 to the two π‘₯. This gives us π‘₯ minus three over two 𝑒 to the two π‘₯ minus one-half times one-half 𝑒 to the two π‘₯. And now, since this is the last integration we will perform, we mustn’t forget to add on our constant of integration or 𝑐, giving us plus 𝑐 on the end. Next, we have one-half times one-half 𝑒 to the two π‘₯. So we can write this as one-quarter 𝑒 to the two π‘₯. And then, we notice that we have some 𝑒 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 𝑒 the two π‘₯ minus three over two 𝑒 to the two π‘₯ minus one-quarter 𝑒 to the two π‘₯ plus 𝑐. When we simplify this, we find our solution which is that the integral of π‘₯ minus three 𝑒 to the two π‘₯ with respect to π‘₯ is equal to π‘₯ over two times 𝑒 to the two π‘₯ minus seven over four times 𝑒 to the two π‘₯ plus 𝑐.

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