Question Video: Finding the Integration of a Function Involving Trigonometric Functions by Distributing the Division

Evaluate ∫_(1) ^(2) π‘₯Β²(π‘₯Β³ βˆ’ 3)Β² dπ‘₯.


Video Transcript

Evaluate the definite integral between one and two of π‘₯ squared times π‘₯ cubed minus three cubed with respect to π‘₯.

This is not a polynomial that’s nice to integrate using our standard rules of finding the antiderivative. And we certainly don’t want to distribute our parentheses and find the antiderivative for each term. Instead, we spot that our integral is set up in this form. It’s the definite integral between some limits of π‘Ž and 𝑏 of some function of 𝑔 of π‘₯ times the derivative of the inner part of that composite function. Here, 𝑔 of π‘₯, the inner part of our composite function, is π‘₯ cubed minus three.

And then, we have a scalar multiple of its derivative here. So, we use integration by substitution to evaluate the integral. This says that if 𝑔 prime, the derivative of 𝑔, is continuous on some closed interval π‘Ž to 𝑏, and 𝑓 is continuous on the range of 𝑒, which is our function 𝑔 of π‘₯. Then, then the definite integral is equal to the definite integral between 𝑔 of π‘Ž and 𝑔 of 𝑏 of 𝑓 of 𝑒 with respect to 𝑒. So, we let 𝑒 be equal to the function that we defined as 𝑔 of π‘₯. It’s the function inside a function whose derivative also appears.

So, we’re going to let 𝑒 be equal to π‘₯ cubed minus three. Now, this is great as when we differentiate 𝑒 with respect to π‘₯, we see that d𝑒 by dπ‘₯ equals three π‘₯ squared. And in integration by substitution, we think of d𝑒 and dπ‘₯ as differentials. And we can alternatively write this as d𝑒 equals three π‘₯ squared dπ‘₯. Notice that whilst d𝑒 by dπ‘₯ is definitely not a fraction, we do treat it a little like one in this process. We divide both sides by three. And we see that a third d𝑒 equals π‘₯ squared dπ‘₯.

So, let’s look back to our original integral. We now see that we can replace π‘₯ squared dπ‘₯ with a third d𝑒. And we can replace π‘₯ cubed minus three with 𝑒. But what do we do with our limits of one and two? Well, we need to replace them with 𝑔 of one and 𝑔 of two. Well, we go back to our original substitution. We said that 𝑒 is equal to π‘₯ cubed minus three, and our lower limit is when π‘₯ is equal to one. So, that’s when 𝑒 is equal to one cubed minus three, which is equal to negative two.

Our upper limit is when π‘₯ is equal to two. So, at this stage, 𝑒 is equal to two cubed minus three, which is of course five. We can now replace each part of our integral with the various substitution. And we see that we’re going to need to work out the definite integral between negative two and five of a third 𝑒 cubed with respect to 𝑒. Remember, we can take any constant factors outside of the integral and focus on integrating 𝑒 cubed.

And then, we recall that we can integrate a polynomial term whose exponent is not equal to negative one by adding one to that exponent and then dividing by that value. So, the integral of 𝑒 cubed is 𝑒 to the fourth power divided by four. And this means our integral is equal to a third times 𝑒 to the fourth power divided by four evaluated between negative two and five. We’ll substitute 𝑒 equals five and 𝑒 equals negative two, and find their difference.

In this case, that’s simply a third of five to the fourth power divided by four minus negative two to the fourth power divided by four. That gives us a third of 609 over four. And we can cancel through by three. And we obtain our solution to be equal to 203 over four, or 50.75. And so, the definite integral between one and two of π‘₯ squared times π‘₯ cubed minus three cubed with respect to π‘₯ is 50.75.

In this example, we saw that we should try and choose 𝑒 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. If that’s not possible though, we try choosing 𝑒 to be a more complicated part of the integrand. This might be the inner function in a composite function or similar.

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