Question Video: Evaluating the Definite Integration of a Quadratic Function Mathematics

Video Transcript

Evaluate the definite integral from negative two to one of 𝑢 plus two times 𝑢 plus one with respect to 𝑢.

In this question, we’re asked to evaluate the definite integral of the product of two linear functions. We know this will give us a quadratic. And we know the definite integral represents the area under a curve. But we don’t know how to directly find the area under a quadratic curve. So we’ll have to do this by using the fundamental theorem of calculus.

We recall the following part of the fundamental theorem of calculus. If lowercase 𝑓 is continuous on a closed interval from 𝑎 to 𝑏 and capital 𝐹 prime of 𝑢 is equal to lowercase 𝑓 of 𝑢, then the definite integral from 𝑎 to 𝑏 of lowercase 𝑓 of 𝑢 with respect to 𝑢 is equal to capital 𝐹 of 𝑏 minus capital 𝐹 of 𝑎. In other words, if our integrand is continuous on the interval of integration and we can find an antiderivative of our integrand, then we can evaluate our definite integral by evaluating our antiderivative.

The first thing we’ll need to do is check that we can use the fundamental theorem of calculus on this integral. We see the lower limit of integration is negative two and the upper limit is one. So we’ll set 𝑎 equal to negative two and 𝑏 equal to one. Next, we’ll need to set our integrand to be our function lowercase 𝑓 of 𝑢. To use the fundamental theorem of calculus, we need to check that the integrand is continuous on the interval of integration. In this case, that’s the closed interval from negative two to one.

In this case, we can see that our integrand is the product of two linear functions. And we know this will give us a quadratic. And a quadratic is an example of a polynomial. And we know all polynomials are continuous over all the real numbers. Therefore, it will be continuous on any closed interval. This means all we need to do now to evaluate this definite integral is find the antiderivative of our integrand.

And the easiest way to do this is to remember the definite integral of 𝑢 plus two times 𝑢 plus one with respect to 𝑢 will give us the general antiderivative of this function. So we just need to evaluate this definite integral. We’ll do this by distributing our parentheses. Distributing our parenthesis and simplifying, we get the integral of 𝑢 squared plus three 𝑢 plus two with respect to 𝑢.

And now, this is just the integral of a polynomial. And we can do this term by term by using the power rule for integration. We want to add one to our exponents of 𝑢 and then divide by this new exponent. This gives us 𝑢 cubed over three plus three 𝑢 squared over two plus two 𝑢. And remember, we need to add a constant of integration 𝐶. And this will be an antiderivative of our integrand for any value of 𝐶. But to apply the fundamental theorem of calculus, we can use any antiderivative. So we can pick our value of 𝐶. We’ll choose 𝐶 equal to zero.

We’re now ready to evaluate our integral by using the fundamental theorem of calculus. First, we rewrite our integrand as 𝑢 squared plus three 𝑢 plus two. Next, by using the power rule of integration, we were able to find the antiderivative 𝑢 cubed over three plus three 𝑢 squared over two plus two 𝑢. And we know by using the fundamental theorem of calculus, we can evaluate our definite integral by evaluating our antiderivative at the upper and lower limits of integration. So we just need to evaluate our antiderivative at one and then subtract our antiderivative evaluated at negative two.

We’ll start by substituting in 𝑢 is equal to one. This gives us one cubed over three plus three times one squared over two plus two times one. And we can just calculate this expression. It’s equal to 23 over six. Then from this, we need to subtract our antiderivative evaluated at negative two. Substituting in 𝑢 is equal to negative two, we get negative two cubed over three plus three times negative two squared over two plus two times negative two. And if we evaluate this expression, we get negative two over three.

So this all simplified to give us 23 over six minus negative two over three. And we can calculate this is equal to nine over two, which is our final answer. Therefore, by using the fundamental theorem of calculus and the power rule for integration, we were able to show the definite integral from negative two to one of 𝑢 plus two times 𝑢 plus one with respect to 𝑢 is equal to nine over two.