Question Video: Evaluating the Definite Integral of a Constant Function Mathematics • Higher Education

Evaluate ∫_(βˆ’3)^(3) βˆ’5𝑒 dπ‘₯.

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Video Transcript

Evaluate the definite integral from negative three to three of negative five 𝑒 with respect to π‘₯.

In this question, we’re asked to evaluate a definite integral. And we’ll do this by using what we know from the fundamental theorem of calculus. So we’ll start by recalling the fundamental theorem of calculus. In fact, we’ll only recall the part about evaluating definite integrals. We know if lowercase 𝑓 is continuous on a closed interval from π‘Ž to 𝑏 and capital 𝐹 prime of π‘₯ is equal to lowercase 𝑓 of π‘₯, then the definite integral from π‘Ž to 𝑏 of lowercase 𝑓 of π‘₯ with respect to π‘₯ is equal to capital 𝐹 evaluated at 𝑏 minus capital 𝐹 evaluated at π‘Ž.

In other words, if our integrand is continuous on the interval of integration, then we can evaluate the definite integral by finding an antiderivative of our integrand. 𝑓 prime of π‘₯ is our antiderivative. We know a lot of different methods of finding antiderivatives. But before we do this, we need to check that our integrand is continuous on the interval of integration. First, for the definite integral given to us in the question, our lower limit of integration is negative three and our upper limit is three. So we set π‘Ž equal to negative three and 𝑏 equal to three.

So we need to show our integrand of negative five 𝑒 is continuous on the closed interval from negative three to three. In this case, our integrand is a constant function. And we know that constant functions are continuous for all real values. So in particular, it will be continuous on the closed interval from negative three to three. This means we’re allowed to evaluate this definite integral by using the fundamental theorem of calculus. We just need to find our antiderivative, capital 𝐹 of π‘₯.

And in fact, we know several different ways of finding this antiderivative. For example, we know the derivative of negative five 𝑒π‘₯ with respect to π‘₯ is just equal to negative five 𝑒. We know this by either using the power rule for differentiation or by noticing that this is a linear function. This means that negative five 𝑒π‘₯ is an example of our antiderivative.

This is not the only method we have for finding antiderivatives, however. We can also use what we know about indefinite integrals. For example, by using the power rule for integration, we know the indefinite integral of negative five 𝑒 with respect to π‘₯ is equal to negative five 𝑒π‘₯ plus a constant of integration 𝐢. And this gives us antiderivatives for any value of 𝐢. We call this the general antiderivative.

We can use whichever method we prefer. And in fact, we can use any value of 𝐢 in our antiderivative, but it’s usually easiest to choose 𝐢 is equal to zero. So we’ll choose our antiderivative capital 𝐹 of π‘₯ to be negative five 𝑒π‘₯. We’re now ready to apply the fundamental theorem of calculus. But before we start substituting 𝑏 and π‘Ž into our antiderivative, there’s one piece of notation we’ll go over. We could just directly substitute our limits of integration into our antiderivative. However, we normally write this using the following notation. We write our antiderivative inside of square brackets and then have our upper and lower limits of integration on the outside. This is just used to help us keep things neat and tidy.

We’re now ready to evaluate our antiderivative at the limits of integration. Remember, we need capital 𝐹 of 𝑏 minus capital 𝐹 of π‘Ž. Doing this gives us negative five 𝑒 times three minus negative five 𝑒 times negative three. And we can then simplify. Negative five times three in our first term is negative 15. And in our second term, negative one times negative five multiplied by negative three simplifies to give us negative 15. So our definite integral simplified to give us negative 15𝑒 minus 15𝑒, which is equal to negative 30𝑒. And this is our final answer.

Therefore, by using the fundamental theorem of calculus, we were able to show the definite integral from negative three to three of negative five 𝑒 with respect to π‘₯ is equal to negative 30𝑒.

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