Video Transcript
Evaluate the integral between zero
and two of two sin 𝑥 minus three 𝑒 to the 𝑥 with respect to 𝑥.
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 𝑎 and 𝑏 and capital 𝐹 is
any antiderivative of lowercase 𝑓. Then the integral between 𝑎 and 𝑏
of lowercase 𝑓 of 𝑥 with respect to 𝑥 is equal to capital 𝐹 of 𝑏 minus capital
𝐹 of 𝑎. 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 𝑎
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 𝑥 is
negative two cos 𝑥. And the antiderivative of negative
three 𝑒 to the 𝑥 is negative three 𝑒 to the 𝑥. 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.
