Video: Evaluating the Definite Integral of a Function Containing a Root Using Integration by Substitution

Evaluate ∫_3^5 π‘₯√(5π‘₯Β² + 3) dπ‘₯ to the nearest thousandth.

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Video Transcript

Evaluate the definite integral from three to five of π‘₯ multiplied by the square root of five π‘₯ squared plus three with respect to π‘₯ to the nearest thousandth.

Let’s look at the function we’ve been asked to integrate more closely. It consists of a polynomial function, π‘₯, multiplied by the square root of another polynomial function, five π‘₯ squared plus three. We have a variety of integration methods available to us, methods such as integration by substitution or integration by parts. And in order to decide which method is appropriate, we need to consider what form the integrand is in. Well, if we look closely, we can see that the derivative with respect to π‘₯ of the function under the square root, that’s five π‘₯ squared plus three, is 10π‘₯, which is a scalar multiple of the function we’re multiplying by, π‘₯. And so our integrand is of the form 𝑓 of 𝑔 of π‘₯, a composite function, multiplied by 𝑔 prime of π‘₯, the derivative of the inner function, subject to multiplication by a constant.

This tells us that we could answer this question using the method of integration by substitution. To do this, we introduce a new variable. So we’re going to let 𝑒 equal five π‘₯ squared plus three. That’s the inner function in our composite function. And we want to change everything in this integral from something in terms of π‘₯ to something in terms of 𝑒. We’ve already said that the derivative of five π‘₯ squared plus three, so that’s now d𝑒 by dπ‘₯, is equal to 10π‘₯. Now, d𝑒 by dπ‘₯ is absolutely not a fraction. But we can treat it a little like one when we’re performing this method. So it’s equivalent to say that d𝑒 is equal to 10π‘₯ dπ‘₯. All we can say that one-tenth of d𝑒 is equal to π‘₯ dπ‘₯. And the reason this is useful is because in our original integral, we have π‘₯ dπ‘₯, which we can now replace with one-tenth d𝑒.

So, let’s now make this substitution and change our integral to be in terms of 𝑒. Firstly, the square root of five π‘₯ squared plus three can be written as the square root of 𝑒. Then, π‘₯dπ‘₯ can be replaced with one-tenth d𝑒. But we’re not done. There’s one more important step. Because this is a definite integral, we must remember to change its limits from limits in terms of π‘₯ to limits in terms of 𝑒. For our upper limit, we substitute π‘₯ equals five into the equation connecting π‘₯ and 𝑒. And we get 𝑒 equals five multiplied by five squared plus three, which is 128. For our lower limit, we substitute π‘₯ equals three into the same equation, giving 𝑒 equals five multiplied by three squared plus three, which is 48. Our integral has now become the definite integral from 48 to 128 of the square root of 𝑒 one-tenth d𝑒. And so we’ve changed it to be fully in terms of 𝑒.

We can rewrite the square root of 𝑒 as 𝑒 to the power of one-half. And if it helps, we can also bring that factor of one-tenth out the front of the integral. So we have one-tenth, the integral from 48 to 128, of 𝑒 to the power of one-half with respect to 𝑒. To integrate a power of 𝑒, where the power is not equal to negative one, we recall that we increase the power or the exponent by one. So that gives 𝑒 to the power of three over two. And then, we divide by the new exponent. Dividing by three over two is equivalent to multiplying by two over three. So our integral is equal to one-tenth of two-thirds 𝑒 to the power of three over two evaluated between 48 and 128. And because we changed the limits in this definite integral, there’s no need to reverse our substitution and go back to an answer in terms of π‘₯.

Our constants of one-tenth and two-thirds can be combined to give an overall constant of one fifteenth. And we then substitute our limits, giving one fifteenth of 128 to the power of three over two minus 48 to the power of three over two. We can use a calculator to evaluate this. And it gives 74.37339 continuing. Finally, we recall that the question asks us to give our answer to the nearest thousandth. So rounding this value, we obtain 74.373.

Using the method of integration by substitution then, we found that the definite integral from three to five of π‘₯ multiplied by the square root of five π‘₯ squared plus three with respect to π‘₯ is equal to 74.373 to the nearest thousandth.

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