Question Video: Evaluating a Definite Integral of a Product by Using a Subsitution Mathematics • Higher Education

Evaluate ∫_(0) ^(3) 4π‘₯Β² (sin ((πœ‹/6) π‘₯Β³)) dπ‘₯.

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Video Transcript

Evaluate the integral from zero to three of four π‘₯ squared times the sin of πœ‹ by six times π‘₯ cubed with respect to π‘₯.

We’re asked to evaluate the definite integral of the product of two functions. There’s a few different ways we might be tempted to approach this. For example, we might be tempted to approach this by using integration by parts. We can see if we differentiated four π‘₯ squared twice, we would end up with a constant. So we might try integration by parts. However, we would then need to integrate a function of the form sin π‘₯ cubed, which we don’t know how to do. However, there is a simpler method. We need to notice the derivative of πœ‹ by six π‘₯ cubed with respect to π‘₯ will be a constant multiple of four π‘₯ squared. And if our integrand is in this form, we can try using integration by substitution.

We’ll want to use the substitution 𝑒 is equal to πœ‹ by six times π‘₯ cubed. We’ll need to differentiate both sides of this with respect to π‘₯. And we can do this by using the power rule for differentiation. This gives us d𝑒 by dπ‘₯ is three times πœ‹ by six multiplied by π‘₯ squared. And of course, we can simplify. Three times πœ‹ by six is equal to πœ‹ by two, so we have d𝑒 by dπ‘₯ is πœ‹ by two times π‘₯ squared. And this is almost exactly what appears in our integrand. To make this easier to work with, we’ll divide both sides of this equation through by πœ‹ by two. And on the left-hand side of this equation, instead of dividing by πœ‹ by two, we’ll multiply by the reciprocal of πœ‹ by two. This gives us two over πœ‹ times d𝑒 by dπ‘₯ is equal to π‘₯ squared.

Finally, we want an expression for four π‘₯ squared, so we’ll multiply both sides of this equation through by four. And doing this gives us eight over πœ‹ d𝑒 by dπ‘₯ is equal to four π‘₯ squared. And it’s worth reiterating at this point d𝑒 by dπ‘₯ is not a fraction; however, when we’re using integration by substitution, it can help to think of it a little bit like a fraction. And this gives us the equivalent statement in terms of differentials. Eight over πœ‹ d𝑒 is equal to four π‘₯ squared dπ‘₯. We’re now almost ready to use our substitution. However, remember, we’re dealing with a definite integral, so we need to find the new limits of integration.

And to do this, we need to substitute these values of π‘₯ into our expression for 𝑒. To find our new upper limit of integration, we substitute π‘₯ is equal to three. This gives us 𝑒 is equal to πœ‹ by six times three cubed. And if we calculate this, we get 27πœ‹ by six, which simplifies to give us nine πœ‹ by two. And we can do the same to find our new lower limit of integration. We substitute in π‘₯ is equal to zero, giving us πœ‹ by six times zero cubed which is, of course, equal to zero. We’re now ready to evaluate our integral by using our substitution. First, our new lower limit of integration will be zero. And our new upper limit of integration will be nine πœ‹ by two.

Next, recall we showed in terms of differentials four π‘₯ squared dπ‘₯ is equivalent to eight over πœ‹ d𝑒, so we can just replace this with eight over πœ‹ d𝑒. Finally, we can use our substitution 𝑒 is equal to πœ‹ by six π‘₯ cubed. And this gives us the integral from zero to nine πœ‹ by two of the sin of 𝑒 times eight over πœ‹ with respect to 𝑒. And at this point, we can just evaluate this integral. Eight over πœ‹ is a constant, and we know the integral of the sin of 𝑒 with respect to 𝑒 will be negative the cos of 𝑒. So by taking our constant factor of eight over πœ‹ outside of our integral and then integrating, we get eight over πœ‹ times negative the cos of 𝑒 evaluated at the limits of integration. 𝑒 is equal to zero, and 𝑒 is equal to nine πœ‹ by two.

And all that’s left to do is evaluate this at the limits of integration. Evaluating this at the limits of integration, we get eight over πœ‹ times negative the cos of nine πœ‹ by two minus negative the cos of zero. And at this point, we can just evaluate this expression. First, the cos of nine πœ‹ by two is equal to zero. Next, the cos of zero is equal to one. And of course, if we’re subtracting negative the cos of zero, then we’re in fact just adding one. So this simplifies to give us eight over πœ‹ times zero plus one, which is just equal to eight over πœ‹.

Therefore, by using the substitution 𝑒 is equal to πœ‹ by six times π‘₯ cubed, we were able to show the integral from zero to three of four π‘₯ squared times the sin of πœ‹ by six multiplied by π‘₯ cubed with respect to π‘₯ is equal to eight over πœ‹.

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