Video: AP Calculus AB Exam 1 • Section I • Part A • Question 13

Calculate ∫_(−1) ^(𝑒 − 2) (5/(𝑥 + 2)) d𝑥.

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Video Transcript

Calculate the integral from negative one to 𝑒 minus two of five over 𝑥 plus two with respect to 𝑥.

In this question, we’re looking to evaluate an integral between the limits of negative one and 𝑒 minus two. We’ll begin then by integrating five over 𝑥 plus two with respect to 𝑥. Now whilst it might look like we need to perform a substitution to be able to complete this integral, there is in fact a standard result that we can quote. We can integrate one over 𝑎𝑥 plus 𝑏 with respect to 𝑥 for constants 𝑎 and 𝑏. And we get one over 𝑎 times ln of 𝑎𝑥 plus 𝑏 plus the constant of integration 𝑐. Remember we can also take a constant multiple outside of the integral and that allows us to focus on integrating the function in 𝑥.

So we rewrite our integral as five times the integral of one over 𝑥 plus two with respect to 𝑥. And now we can see that the integral of one over 𝑥 plus two with respect to 𝑥 is ln of 𝑥 plus two. And since we’re evaluating this between the limits of negative one and 𝑒 minus two, we don’t need to worry about that constant of integration.

We now substitute 𝑒 minus two and negative one into our integral. And it becomes five times ln of 𝑒 minus two plus two minus ln of negative one plus two. This simplifies to five times ln of 𝑒 minus ln of one. But remember ln of 𝑒 is simply one and ln of one is zero. So we get five times one which is equal to five.

And so, we see that the integral between the limits of negative one and 𝑒 minus two of five over 𝑥 plus two with respect to 𝑥 is five.

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