Video: Evaluating the Definite Integration of a Function Involving Exponential and Trigonometric Functions

Evaluate ∫_(0) ^(2) (2 sin π‘₯ βˆ’ 3𝑒^π‘₯) dπ‘₯.


Video Transcript

Evaluate the integral between zero and two of two sin π‘₯ minus three 𝑒 to the π‘₯ with respect to π‘₯.

To answer this question, we’re going to be using the second part of the fundamental theorem of calculus. This tells us that if lowercase 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏 and capital 𝐹 is any antiderivative of lowercase 𝑓. Then the integral between π‘Ž and 𝑏 of lowercase 𝑓 of π‘₯ with respect to π‘₯ is equal to capital 𝐹 of 𝑏 minus capital 𝐹 of π‘Ž. Looking back at our question, the first thing we might notice is that the function lowercase 𝑓, which is our integrand, consists of two different terms. The first term involves the trigonometric function sine. And the second involves the exponential 𝑒. Now, we should be familiar with the fact that both trigonometric and exponential functions of this form are continuous over the entire set of real numbers. We have therefore fulfilled the criteria that our function 𝑓 must be continuous over the closed interval between π‘Ž and 𝑏, which in our case is the closed interval between zero and two.

Now, given that we do have two terms, we might find that our working is clearer if we split these up into separate integrals. We would do this like so, remembering to keep the limits of integration the same across both terms. We can now evaluate each of these integrals separately. The antiderivative of two sin π‘₯ is negative two cos π‘₯. And the antiderivative of negative three 𝑒 to the π‘₯ is negative three 𝑒 to the π‘₯. Of course, remember, we can ignore the constant of integration in both cases since we’re working with definite integrals.

Here, we note that we’ve expressed our antiderivative in brackets, with the limits of integration being carried over to the right-hand bracket in both cases. Given that both of these brackets have the same limits of integration, we can simply combine them. Now, you might notice that we could’ve moved directly from our original integral to this step, by treating each of the terms individually. Instead of splitting our integral into two and then recombining, we would simply have found the antiderivative of each of our terms. If you’re not sure, however, there’s no harm in writing out the method in full.

To move forward with our question, we now substitute in the limits of our integration, which are zero and two. We then reach the following expression. With our first set of parentheses, there’re no simplifications necessary. So we can just leave this. For the second set of parentheses, we might recall that cos of zero is equal to one. So negative two cos zero is equal to negative two. Alongside this, 𝑒 to the power of zero is also one. So negative three 𝑒 to the power of zero is negative three. Our second set of parentheses therefore becomes negative two minus three, which is negative five. But for our final answer, we’re subtracting this. So we’re left with a positive five. And we’ve now reached our final answer. The definite integral given in the question evaluates to negative two cos of two minus three 𝑒 squared plus five.

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