### Video Transcript

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

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

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. π 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 equal to negative three and π equal to three.

So we need to show our integrand of negative five π 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 weβre allowed to evaluate this definite integral by using the fundamental theorem of calculus. We just need to find our antiderivative, capital πΉ of π₯.

And in fact, we know several different ways of finding this antiderivative. For example, we know the derivative of negative five ππ₯ with respect to π₯ is just equal to negative five π. 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 ππ₯ 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 π with respect to π₯ is equal to negative five ππ₯ plus a constant of integration πΆ. And this gives us antiderivatives for any value of πΆ. We call this the general antiderivative.

We can use whichever method we prefer. And in fact, we can use any value of πΆ in our antiderivative, but itβs usually easiest to choose πΆ is equal to zero. So weβll choose our antiderivative capital πΉ of π₯ to be negative five ππ₯. Weβre now ready to apply the fundamental theorem of calculus. But before we start substituting π and π into our antiderivative, thereβs one piece of notation weβll 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.

Weβre now ready to evaluate our antiderivative at the limits of integration. Remember, we need capital πΉ of π minus capital πΉ of π. Doing this gives us negative five π times three minus negative five π 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 15π minus 15π, which is equal to negative 30π. 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 π with respect to π₯ is equal to negative 30π.