Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution

Determine ∫ ((48 − 6𝑥)/^5√(16 − 2𝑥)) d𝑥.


Video Transcript

Determine the integral of 48 minus six 𝑥 over the fifth root of 16 minus two 𝑥 d𝑥.

It’s not instantly obvious how we need to evaluate this integral. However, if we look carefully, we see that the numerator is a scalar multiple of the inner function on the denominator such that 48 minus six 𝑥 is three times 16 minus two 𝑥. This is a hint to us that we’re going to need to use integration by substitution to evaluate this indefinite integral.

We’re going to let 𝑢 be equal to 16 minus two 𝑥. We’ve chosen this part for our substitution as 16 minus two 𝑥 is the inner function in a composite function. We differentiate 𝑢 with respect to 𝑥, and we see that d𝑢 by d𝑥 is equal to negative two. Now, remember, d𝑢 by d𝑥 is not a fraction, but we treat it a little like one when performing integration by substitution. And we can see that this is equivalent to saying negative one-half d𝑢 is equal to d𝑥.

Let’s substitute what we now have into our original integral. We saw that 48 minus six 𝑥 is equal to three times 16 minus two 𝑥. So, the numerator becomes three 𝑢. The denominator becomes the fifth root of 𝑢. And we replace d𝑥 with negative a half d𝑢. Let’s take out a factor of negative three over two. And we’ll write our denominator as 𝑢 to the fifth power. We’re dividing 𝑢 to the power of one by 𝑢 to the power of one-fifth.

So, we subtract one-fifth from one, and we’re left with 𝑢 to the four-fifths. The antiderivative of 𝑢 to the four-fifths is 𝑢 to the nine-fifths divided by nine-fifths. That’s the same as five-ninths times 𝑢 to the nine-fifths. And, remember, this is an indefinite integral, so we add that constant of integration 𝑐.

When we distribute the parentheses, we have negative five-sixths times 𝑢 to the nine-fifths plus capital 𝐶. Since our original constant has been multiplied by negative three over two, and since we were evaluating an integral in terms of 𝑥, we must replace 𝑢 with 16 minus two 𝑥. And we see that our integral is negative five-sixths times 16 minus two 𝑥 to the nine-fifths plus 𝐶.

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