Video: Evaluating the Definite Integral of a Function Using the Power Rule for Integration

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

06:04

Video Transcript

Evaluate the integral from one to two of two minus π‘Ÿ all cubed with respect to π‘Ÿ.

The question is asking us to evaluate the definite integral of a linear function all cubed. And there’s actually two different ways of evaluating this integral. First, we can rewrite our integrand by using the binomial expansion formula. We recall our binomial expansion formula tells us for a positive integer 𝑛, π‘Ž plus 𝑏 all raised to the 𝑛th power is equal to π‘Ž to the 𝑛th power plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one times 𝑏 plus 𝑛 choose two π‘Ž to the power of 𝑛 minus two times 𝑏 squared. And we add terms in this fashion all the way up to 𝑏 to the power of 𝑛.

In our case, our value of 𝑛 is three. So by using 𝑛 is equal to three, we get π‘Ž plus 𝑏 all cubed is equal to π‘Ž cubed plus three choose one π‘Ž squared 𝑏 plus three choose two π‘Žπ‘ squared plus 𝑏 cubed. And we can simplify this since both three choose one and three choose two are equal to three. So this gives us π‘Ž cubed plus three π‘Ž squared 𝑏 plus three π‘Žπ‘ squared plus 𝑏 cubed. We now want to use this to rewrite our integrand. Our value of π‘Ž will be two and our value of 𝑏 will be negative π‘Ÿ. So to find an expression for two minus π‘Ÿ all cubed, we substitute π‘Ž is equal to two and 𝑏 is equal to negative π‘Ÿ into our expression.

We get two minus π‘Ÿ all cubed is equal to two cubed plus three times two squared multiplied by negative π‘Ÿ plus three times two multiplied by negative π‘Ÿ squared plus negative π‘Ÿ cubed. And if we simplify each term, we get eight minus 12π‘Ÿ plus six π‘Ÿ squared minus π‘Ÿ cubed. And we can see that this is now an expression which we can integrate. We’re now ready to use this to evaluate our integral. First, we’ll rewrite our integrand by using the expression we found using the binomial expansion formula. Now, we see we can evaluate this integral term by term by using the power rule for integration. We want to add one to each of our exponents of π‘Ÿ and then divide by this new exponent of π‘Ÿ.

Using this on each term and simplifying our answers, we get eight π‘Ÿ minus six π‘Ÿ squared plus two π‘Ÿ cubed minus π‘Ÿ to the fourth power over π‘Ÿ [four] evaluated at the limits of our integral one and two. The last thing we need to do is evaluate this at the limits of our integral. Doing this, we get eight times two minus six times two squared plus two times two cubed minus two to the fourth power over four minus eight times one minus six times one squared plus two times one cubed minus one to the fourth power over four. And we can just calculate this expression. We get four minus 15 over four, which is equal to one-quarter. So this is the first way we could’ve evaluated this integral. However, there is a second method we could’ve used.

So we’ll clear some space and go through this second method we could’ve used. The second method we could’ve used is integration by substitution. We notice we’re asked to integrate a composite function. And if we set 𝑒 to be our inner function two minus π‘Ÿ, then we see the derivative of 𝑒 with respect to π‘Ÿ is equal to negative one. And this is just a constant. So we have a trick to put this into our integral. We’ll multiply our integrand by negative one and then multiply our entire integral by negative one. This doesn’t change the value of our integral. However, it rewrites it in a form which we should recognize.

We see this is now in the form negative one multiplied by a definite integral of the form 𝑔 prime of π‘Ÿ times 𝑓 of 𝑔 of π‘Ÿ with respect to π‘Ÿ. This means we can use integration by substitution. So let’s do that by using 𝑒 is equal two minus π‘Ÿ. We want to rewrite this integral so every part is in terms of 𝑒. We’ve already shown that when 𝑒 is equal to two minus π‘Ÿ, d𝑒 by dπ‘Ÿ is equal to negative one. Now remember, d𝑒 by dπ‘Ÿ is not a fraction. However, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑒 is equal to negative one times dπ‘Ÿ.

We’re now ready to try and evaluate our integral by using our substitution. First, we’ll multiply the whole integral and our integrand by negative one. This gives us negative one times the integral from one to two of negative one times two minus π‘Ÿ all cubed with respect to π‘Ÿ. Next, we’re using the substitution 𝑒 is equal to two minus π‘Ÿ. So we’ll replace our inner function with 𝑒. And we showed for the substitution, 𝑒 is equal to two minus π‘Ÿ, negative one times dπ‘Ÿ is equal to d𝑒. We now have negative one times the integral of 𝑒 cubed d𝑒. There’s just one more thing we need to do. We need to find the new limits of our integral.

Our old limits were in terms of π‘Ÿ. So these tell us the values of π‘Ÿ. We need to find the values of 𝑒. Let’s start with the new upper limit of our integral. When π‘Ÿ is equal to two, we get that 𝑒 is equal to two minus two, which is equal to zero. Now, let’s find the lower limit of our integral. When π‘Ÿ is equal to one, we get 𝑒 is equal to two minus one, which is equal to one. So we now have negative one times the integral from one to zero of 𝑒 cubed with respect to 𝑒. Now, we can actually just evaluate this integral as it is. However, we’ll rewrite it by using one of our rules for definite integrals, the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ is equal to negative one times the integral from 𝑏 to π‘Ž of 𝑓 of π‘₯ with respect to π‘₯.

This means we can remove our factor of negative one and then switch the limits of our integral around. So we now just have to evaluate the integral from zero to one of 𝑒 cubed with respect to 𝑒. And we can just evaluate this by using the power rule for integration. We get 𝑒 to the fourth power over four evaluated at the limits of our integral zero and one. Finally, we just evaluate this at the limits of our integral. We get one to the fourth power over four minus zero to the fourth power over four which we can calculate is equal to one-quarter.

Therefore, we’ve shown two different methods of evaluating the integral from one to two of two minus π‘Ÿ all cubed with respect π‘Ÿ. We got that this was equal to one-quarter.

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