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

Video Transcript

Let 𝑓 of 𝑥 be equal to six 𝑥 squared plus one. Evaluate the definite integral of the function 𝑓 from 𝑥 is equal to two to 𝑥 is equal to three.

First, we see that the question is asking us to evaluate the definite integral of the function 𝑓 within the limits of 𝑥 is equal to two and 𝑥 is equal to three, with respect to 𝑥. We can also see from the question that the function 𝑓 of 𝑥 is defined to equal six 𝑥 squared plus one. So, we can rewrite the 𝑓 of 𝑥 in our integral as just six 𝑥 squared plus one.

Now, we know that for an antiderivative, capital 𝑓 of 𝑥 of the function lowercase 𝑓 of 𝑥, we will have that the definite integral from 𝑎 to 𝑏 of lowercase 𝑓 of 𝑥 with respect to 𝑥 is equal to capital 𝑓 of 𝑏 minus capital 𝑓 of 𝑎. Therefore, we could use this to solve the integral in our question.

If we had the antiderivative of six 𝑥 squared plus one, then evaluating this antiderivative at three and subtracting the antiderivative evaluated at two would give us the answer to our question. We can find the antiderivative of 𝑓 of 𝑥 by using the power rule for integration which says that if 𝑛 is not equal to negative one, then the integral of 𝑘 multiplied by 𝑥 to the nth power with respect to 𝑥 is equal to 𝑘 multiplied by 𝑥 to the 𝑛 plus oneth power divided by 𝑛 plus one plus a constant of integration. It’s worth noting, at this point, that since we’re dealing with a definite integral, the constants of integration will always cancel out in our calculation. So, we can safely leave this out.

Integrating our first term of six 𝑥 squared gives us six 𝑥 to the power of two plus one divided by two plus one, which we can simplify to two 𝑥 cubed. We also know that one is equal to 𝑥 to the zeroth power. Therefore, we can use the power rule for integration to see that the integral of one with respect to 𝑥 is equal to 𝑥 to the one plus zero divided by one plus zero. Which simplifies to just give us 𝑥, meaning that our antiderivative of the function lowercase 𝑓 of 𝑥 is equal to two 𝑥 cubed plus 𝑥.

We are now ready to evaluate this in our limits of two and three. Evaluating the antiderivative of three and subtracting the evaluation of the antiderivative of two gives us two multiplied by three cubed plus three minus two multiplied by two cubed plus two. Which we can calculate to give us an answer of 39. Therefore, what we have shown is that the definite integral from 𝑥 is equal to two to 𝑥 equals three of the function lowercase 𝑓 of 𝑥 is equal to six 𝑥 squared plus one with respect to 𝑥 is equal to 39.