Video Transcript
Given that capital 𝐹 of 𝑥 is equal to the definite integral from the square root of 𝑥 to four 𝑥 of the inverse tan of 𝑡 with respect to 𝑡, find capital 𝐹 prime of 𝑥.
We’re given a function capital 𝐹 of 𝑥, which is defined as a definite integral where the limits of integration are functions in 𝑥. We need to use this definition for capital 𝐹 of 𝑥 to find 𝐹 prime of 𝑥. Since 𝐹 is a function in 𝑥, this will be the derivative of capital 𝐹 of 𝑥 with respect to 𝑥. And therefore, capital 𝐹 prime of 𝑥 will be the derivative of an integral where the limits of integration are functions in 𝑥. This should remind us of the fundamental theorem of calculus.
So we’ll start by recalling the following part of the fundamental theorem of calculus. If lowercase 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and we have 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 will be equal to lowercase 𝑓 evaluated at 𝑥 for all values of 𝑥 in the open interval from 𝑎 to 𝑏. In other words, the fundamental theorem of calculus gives us a way of differentiating an integral where 𝑥 is in the limit of integration.
Of course, this isn’t quite true to the function we’re given in the question. We can see our upper limit of integration is not 𝑥. In fact, it’s a function in 𝑥. We can also see the lower limit of integration is not a constant; it’s also a function in 𝑥. We can get around both of these problems. We’ll start by recalling a slight modification to our fundamental theorem of calculus.
By using our standard statement of the fundamental theorem of calculus and the chain rule, we have if 𝑢 of 𝑥 is a differentiable function, lowercase 𝑓 is continuous on the closed interval from 𝑎 to 𝑏, and capital 𝐹 of 𝑥 is equal to the definite integral from 𝑎 to 𝑢 of 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡, then capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 evaluated at 𝑢 of 𝑥 multiplied by 𝑢 prime of 𝑥, as long as 𝑢 of 𝑥 is in the open interval from 𝑎 to 𝑏.
And we can see this is more similar to the function given to us in the question as both of the limits in our integral are differentiable functions. However, we still don’t have our lower limit of integration as a constant. To fix this, we’re going to need to use one of our rules for definite integrals. We need to recall the following fact about definite integrals. The definite integral from 𝑎 to 𝑏 of lowercase 𝑓 of 𝑡 with respect to 𝑡 is equal to the definite integral from 𝑐 to 𝑏 of lowercase 𝑓 of 𝑡 with respect of 𝑡 plus the definite integral from 𝑎 to 𝑐 of lowercase 𝑓 of 𝑡 with respect to 𝑡. In other words, we’re splitting our definite integral at the constant value of 𝑐.
And to use this, we do need to be careful. We need our integrand lowercase 𝑓 of 𝑡 to be integrable on both domains of integration. Otherwise, this won’t work. We need to be careful when we choose our value of 𝑐. So by splitting the definite integral given to us in the question at the constant 𝑐, we get capital 𝐹 of 𝑥 is equal to the definite integral from 𝑐 to four 𝑥 of the inverse tan of 𝑡 with respect to 𝑡 plus the definite integral from root 𝑥 to 𝑐 of the inverse tan of 𝑡 with respect to 𝑡. And before we carry on, remember, this will only be valid if our integrand is integrable over both intervals of integration, so we should determine a suitable value of 𝑐.
To do this, let’s take a closer look at our integrand, the inverse tan of 𝑡. We know something about this function. We know it’s continuous for all real values of 𝑡. And if it’s continuous for all values of 𝑡, this means it’s integrable on any interval. So in this case, we could actually pick any value of 𝑐, so we’ll just choose the value of zero. However, in this case, it doesn’t matter which value we pick. We’re now almost ready to use our modified version of the fundamental theorem of calculus. However, in our second integral, we notice that our constant is in the upper limit of integration and our function is in the lower limit of integration.
But we can fix this by using another one of our rules for definite integrals. We can switch the upper and lower limit of integration if we multiply our integral by negative one. This gives us the following expression for capital 𝐹 of 𝑥. We’re now ready to use our modified version of the fundamental theorem of calculus to find an expression for capital 𝐹 prime of 𝑥. We want to differentiate each of our definite integrals separately. We’ll start with the first definite integral. 𝑢 of 𝑥 is four 𝑥, lowercase 𝑓 of 𝑡 is the inverse tan of 𝑡, and our value of 𝑎 is equal to zero.
Remember though, to use the fundamental theorem of calculus, we do need to check where our integrand lowercase 𝑓 is continuous. However, in this case, we’ve already shown the inverse tan of 𝑡 is continuous for all real values of 𝑡. So in particular, it will be continuous on any closed interval. This means we’ve justified our use of this modified version of the fundamental theorem of calculus. And therefore, we can use this to evaluate the derivative of our first definite integral. It will be lowercase 𝑓 evaluated at four 𝑥 multiplied by the derivative of four 𝑥 with respect to 𝑥.
We’ll want to do exactly the same to find the derivative of our second definite integral. It’s also worth reiterating here, we do need our function 𝑢 of 𝑥 to be differentiable since we use the chain rule. But we know this is true in this case, and we’ve already shown all the prerequisites are true in our first derivative, so we can just apply the result here. We get lowercase 𝑓 evaluated at root 𝑥 multiplied by the derivative of root 𝑥 with respect to 𝑥. And now we can just start evaluating this expression.
First, to find the lowercase 𝑓 evaluated at four 𝑥, we substitute 𝑡 is equal to four 𝑥 into our integrand. This gives us the inverse tan of four 𝑥. Next, we want to multiply this by the derivative of four 𝑥 with respect to 𝑥. We know by using the power rule for integration, this is four. And we’ll write this at the start of our expression to make it easier. And we can do the same for our second term. We need to substitute root 𝑥 into our integrand. This gives us the inverse tan of root 𝑥.
Finally, by using our laws of exponents and the power rule for differentiation, we know the derivative of root 𝑥 with respect to 𝑥 is one divided by two root 𝑥. And we’ll simplify this expression slightly by writing this factor at the start. And doing this gives us our final answer.
Therefore, by using the chain rule and the fundamental theorem of calculus, we were able to find an expression for capital 𝐹 prime of 𝑥. We were able to show that it’s equal to four times the inverse tan of four 𝑥 minus one over two root 𝑥 times the inverse tan of root 𝑥.
