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

Determine โซ_(โ5) ^(โ2) 2/โ(๐ฅ + 6) d๐ฅ.

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Video Transcript

Determine the definite integral between negative five and negative two of two over the square root of ๐ฅ plus six d๐ฅ.

It may not be instantly obvious how weโre going to evaluate this integral. However, if we look carefully, we can see that the numerator is a scalar multiple of the derivative of the inner function on our denominator. In other words, the derivative of ๐ฅ plus six multiplied by two is equal to the numerator. Thatโs two. And thatโs a hint to us that weโre going to need to use integration by substitution to evaluate our integral.

Remember, in integration by substitution, we introduce a new function. That is ๐ข, which we let ๐ equal to ๐ of ๐ฅ. And we see that the definite integral between ๐ and ๐ of ๐ of ๐ of ๐ฅ times ๐ prime of ๐ฅ with respect to ๐ฅ is equal to the definite integral between ๐ of ๐ and ๐ of ๐ of ๐ of ๐ข with respect to ๐ข. Weโre going to let ๐ข be equal to the inner part of our composite function, thatโs ๐ฅ plus six, so that d๐ข by d๐ฅ is equal to one. And whilst d๐ข by d๐ฅ isnโt a fraction, we are allowed to treat it a little like one when weโre working with integration by substitution. And we can say that d๐ข is equal to d๐ฅ.

And so, we can replace d๐ฅ with d๐ข and ๐ฅ plus six with ๐ข. But weโre going to need to do something with our limits. We use our substitution to redefine them. Our lower limit is when ๐ฅ is equal to negative five. So, ๐ข then is equal to ๐ฅ plus six, which is here negative five plus six, which is of course one. Then, our upper limit is when ๐ฅ is equal to negative two. So, ๐ข is equal to negative two plus six, which is four. And so, our integral is equal to the definite integral between one and four of two over the square root of ๐ข with respect to ๐ข.

Now, in fact, we can write one over the square root of ๐ข as ๐ข to the power of negative one-half. And this next step isnโt entirely necessary, but it can make the process simpler. We recall that we can take any constant factors outside of the integral and focus on integrating the function in ๐ข itself. So, this is equal to two times the definite integral between one and four of ๐ข to the power of negative one-half. Now, when we integrate ๐ข to the power of negative one-half, we add one to the exponent and then divide by that new number. So, ๐ข to the power of negative half becomes ๐ข to the power of one-half divided by one-half, which is the same as two times ๐ข to the power of one-half.

We then replace ๐ข with four and one and find the difference. We get two times two lots of four to the power of one-half minus two lots of one to the power of one-half. Well, four to the power of one-half is two. And one to the power of one-half is one. So, we have two times two times two minus two times one, which is simply equal to four. And weโre done. Since we changed our limits, we donโt need to do anything more. Our definite integral is equal to four.