Video Transcript
Let 𝑦 be equal to the integral from the sin of two 𝑥 to two of the square root of two plus five 𝑡 squared with respect to 𝑡. Use the fundamental theorem of calculus to find 𝑦 prime.
In this question, we’re given a function 𝑦, which is defined as a definite integral. And we can see one of the limits of our integral as a function of 𝑥. We need to use the fundamental theorem of calculus to determine an expression for 𝑦 prime. To find an expression for 𝑦 prime, we first need to work out what 𝑦 is a function of. And to do this, we’ll take a closer look at the definition of 𝑦.
We’re taking a definite integral of the square root of two plus five 𝑡 squared with respect to 𝑡. When we’re calculating a definite integral, we find an antiderivative of our integrand and then substitute the limits of integration into this. So the last step of this process is to evaluate at the limits of integration. This means that 𝑦 is a function in 𝑥. So in this case, 𝑦 prime is d𝑦 by d𝑥.
So to answer this question, we’ll start by recalling the fundamental theorem of calculus. In fact, we’ll just recall the first part. This says, if lowercase 𝑓 is a continuous function on a closed interval from 𝑎 to 𝑏 and capital 𝐹 of 𝑥 is the definite integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡, then capital 𝐹 prime of 𝑥 is equal to lowercase 𝑓 of 𝑥 for all values of 𝑥 in the open interval from 𝑎 to 𝑏. So this gives us a method of differentiating a definite integral where 𝑥 is one of the limits of integration. And the function we’re given in the question is almost in this form. For example, 𝑦 will be capital 𝐹 of 𝑥.
Our integrand, the square root of two plus five 𝑡 squared, will be lowercase 𝑓 of 𝑡. However, we then see a problem. To use the fundamental theorem of calculus, our lower limit should be a constant and our upper limit should be 𝑥. However, in this case, our upper limit is a constant and our lower limit is a function in 𝑥. However, we can fix both of these problems.
First, remember, in definite integrals, we can switch the limits of integration around by multiplying our integrand by negative one. So in our expression for 𝑦, we’ll multiply the entire integral by negative one and switch the limits of integration around. This gives us that 𝑦 is equal to negative the integral from two to the sin of two 𝑥 of the square root of two plus five 𝑡 squared with respect to 𝑡. So now, in the fundamental theorem of calculus, our lower limit is a constant.
However, we still have a function in 𝑥 as our upper limit instead of 𝑥. And in fact, we can solve this problem entirely by using what we know about calculus. To start, we’ll call 𝑢 the sin of two 𝑥, which is the upper limit of integration. So by using this, we now have that 𝑦 is equal to negative the definite integral from two to 𝑢 of the square root of two plus five 𝑡 squared with respect to 𝑡. We want to differentiate this entire expression with respect to 𝑥. This would give us that 𝑦 prime is the derivative with respect to 𝑥 of negative the integral from two to 𝑢 of the square root of two plus five 𝑡 squared with respect to 𝑡. And we can see that this expression is almost in a form we can use the fundamental theorem of calculus on.
There’s now only one small problem. Using the exact same logic we did before, in this definite integral, we’re going to need to evaluate this at the limits of integration. This means that this is not a function in 𝑥. In fact, it’s a function in 𝑢. So 𝑦 is now a function in 𝑢. But remember, 𝑢 is a function in 𝑥. So we’re differentiating 𝑦 of 𝑢 with respect to 𝑥, where 𝑦 is a function in 𝑢 and 𝑢 is a function in 𝑥. We know how to do this by using the chain rule.
The chain rule tells us if 𝑦 is a function in 𝑢 and 𝑢 is a function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. This means by applying the chain rule, we have 𝑦 prime is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. And in fact, we can evaluate both of these expressions. First, let’s write our expression for 𝑦 of 𝑢 into d𝑦 by d𝑢. This gives us the derivative with respect to 𝑢 of negative the integral from two to 𝑢 of the square root of two plus five 𝑡 squared with respect to 𝑡 multiplied by d𝑢 by d𝑥.
And there’s one last thing we need to do to this expression to use the fundamental theorem of calculus. We need to take our factor of negative one outside of our derivative. Now, this is exactly in the form we need to use the fundamental theorem of calculus. We just need to change our variable from 𝑥 to 𝑢. So we’ll rewrite the fundamental theorem of calculus with 𝑢 instead of 𝑥. We see that our upper limit of integration is 𝑢. Our value of 𝑎, which is the lower limit of integration, will be two. Our integrand of the square root of two plus five 𝑡 squared will be lowercase 𝑓 of 𝑡. And 𝑦, as a function in 𝑢, will be capital 𝐹 of 𝑢.
The last thing we need to do to use the fundamental theorem of calculus is check that our integrand, lowercase 𝑓, is continuous. And in this case, our integrand is the square root of a polynomial. It’s the composition of continuous functions. This means it’s continuous across its entire domain. And in this instance, two plus five 𝑡 squared is greater than or equal to zero for all real values of 𝑡. So it’s defined for all values of 𝑡, which means it’s continuous for all values of 𝑡.
Therefore, our integrand is continuous on any closed interval. This means we can use the fundamental theorem of calculus to evaluate this derivative. So we just need to substitute 𝑢 into our integrand. Remember, we also multiply our entire derivative by negative one. This gives us negative the square root of two plus five 𝑢 squared. Remember, we also need to multiply this by d𝑢 by d𝑥. But we know 𝑢 is the sin of two 𝑥. So we’re multiplying this expression by the derivative of the sin of two 𝑥 with respect to 𝑥. And we know that this is two times the cos of two 𝑥.
So this gives us the following expression for 𝑦 prime. It’s equal to negative the square root of two plus five 𝑢 squared multiplied by two times the cos of two 𝑥. And we could leave our answer like this. However, remember, 𝑦 is a function in 𝑥. So 𝑦 prime should be written in terms of 𝑥. And we can do this by using our substitution. 𝑢 is the sin of two 𝑥. So by using the substitution and rearranging, we get our final answer. We get that 𝑦 prime is equal to negative two cos of two 𝑥 multiplied by the square root of two plus five times the sin squared of two 𝑥.
