Video: Using Integration by Substitution to Evaluate a Definite Integral

Calculate ∫_(0) ^(2) (3π‘₯Β³ + 2π‘₯)(3π‘₯⁴ + 4π‘₯Β²) dπ‘₯.

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Video Transcript

Calculate the definite integral from zero to two of three π‘₯ cubed plus two π‘₯ multiplied by three π‘₯ to the fourth power plus four π‘₯ squared with respect to π‘₯.

Now, looking at the integrand, that’s the expression we’ve been asked to integrate, we can see that it is the product of two polynomials. There are, in fact, two methods that we could use to answer this question. So, we’ll look at them both. As neither of these polynomials is raised to a power other than one, it’s actually quite straightforward to just distribute the parentheses to give a new polynomial, which we can then integrate term by term. So, that’s what we’ll do in our first method.

Distributing the parentheses and remembering that we add the exponents when multiplying two polynomial terms together. We have that our integral is equal to the definite integral from zero to two of nine π‘₯ to the seventh power plus 12π‘₯ to the fifth power plus six π‘₯ to the fifth power plus eight π‘₯ cubed with respect to π‘₯. We can then combine the like terms and simplify integrand to nine π‘₯ to the seventh power plus 18π‘₯ to the fifth power plus eight π‘₯ cubed.

We then recall that to integrate the general polynomial term π‘Žπ‘₯ to the power of 𝑛 for real values of π‘Ž and 𝑛 where 𝑛 is not equal to negative one, we increase the exponent by one and divide by the new exponent. Giving π‘Žπ‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus the constant of integration 𝑐 if it’s an indefinite integral.

Although our integral has limits, it’s a definite integral. So, we don’t need to worry about constants of integration here. Applying this rule and integrating term by term then, we have that this integral is equal to nine π‘₯ to the eighth power over eight plus 18π‘₯ to the sixth power over six plus eight π‘₯ to the fourth power over four. And that’s all evaluated between the limits of zero and two.

Before substituting the limits, we can simplify some of the coefficients. In the second term, we have a coefficient of 18 over six, which simplifies to three. And in the third term, we have a coefficient of eight over four, which simplifies to two. So, our expression becomes nine π‘₯ to the eighth power over eight plus three π‘₯ to the sixth power plus two π‘₯ to the fourth power.

Substituting our limits then, we have nine multiplied by by two to the eighth power over eight plus three multiplied by two to the sixth power plus two multiplied by two to the fourth power. And then, we know that when we substitute zero, we will just get zero. So, we’re subtracting zero for the lower limit. Evaluating on a calculator gives 288 plus 192 plus 32, which all sums to 512. So, by distributing the parentheses to give a polynomial and then recalling how to integrate a general polynomial term where the exponent is not equal to negative one, we’ve calculated the value of this definite integral to be 512.

Now, there is another method that we could use. And in order to use this method, we need to spot that there is a connection between the expressions inside each set of parentheses. If we find the derivative of the expression in the second set of parentheses, that’s three π‘₯ to the fourth power plus four π‘₯ squared, we find that it’s equal to 12π‘₯ cubed plus eight π‘₯. And if we factor by four, we then see that this is an exact multiple of the expression inside the first set of parentheses.

This suggests that we could also solve this problem by using integration by substitution. Let’s have a look at what that would look like. We introduce a new variable, which we’ll call 𝑒, to be that part of the expression whose derivative, or a scale multiple of it, also appears in the integrand. So, 𝑒 is equal to three π‘₯ to the fourth power plus four π‘₯ squared. We’ve just seen that the derivative of this expression, so that’s now d𝑒 by dπ‘₯, is equal to 12π‘₯ cubed plus eight π‘₯, which we can write as four multiplied by three π‘₯ cubed plus two π‘₯.

Now, d𝑒 by dπ‘₯ is not a fraction, but we can treat it a little like one. So, it’s equivalent to say that one-quarter d𝑒 is equal to three π‘₯ cubed plus two π‘₯ dπ‘₯. Now, why is this helpful? Well, if we look back at the original integrand, we can see that three π‘₯ to the fourth power plus four π‘₯ squared is what we defined to be 𝑒. And three π‘₯ cubed plus two π‘₯ dπ‘₯ is what we’ve just seen to be equal to one-quarter d𝑒. So, we know how to change everything inside our integral from being in terms of π‘₯ to being in terms of 𝑒.

What we must also remember, though, is that we have a definite integral. So, we also need to change our limits of zero and two from limits in terms of π‘₯ into limits in terms of 𝑒. We do this using the equation we already have connecting π‘₯ and 𝑒. When π‘₯ is equal to zero, 𝑒 is equal to three multiplied by zero to the fourth power plus four multiplied by zero squared, which is equal to zero. And when π‘₯ is equal to two, 𝑒 is equal to three multiplied by two to the fourth power plus four multiplied by two squared, which is equal to 64. So, the limits for our integral in terms of 𝑒 are zero and 64.

Let’s now change this integral to be in terms of 𝑒. We have the definite integral between zero and 64. Three π‘₯ to the fourth power plus four π‘₯ squared becomes 𝑒. And three π‘₯ cubed plus two π‘₯ dπ‘₯ becomes one-quarter d𝑒. We can rewrite that factor of one-quarter in front of 𝑒 or even bring it out to the front of the integral if we wish. And we find that our integral has now become the integral from zero to 64 of one-quarter 𝑒 with respect to 𝑒.

Now, this is just the integral of a polynomial term. So, we can apply the same rule that we already used. We increase the exponent by one to give 𝑒 squared and divide by the new exponent to give one-quarter multiplied by a half 𝑒 squared, evaluated between zero and 64. The coefficient simplifies to one-eighth. And then substituting our limits, we have one-eighth of 64 squared minus one-eighth of zero squared, which is just zero. And then once again, evaluated on a calculator, we find that this integral is equal to 512.

So, by either distributing the parentheses and then integrating term by term or using the method of integration by substitution. We’ve calculated that the definite integral from zero to two of three π‘₯ cubed plus two π‘₯ multiplied by three π‘₯ to the fourth power plus four π‘₯ squared with respect to π‘₯ is equal to 512.

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