The Fundamental Theorem of Calculus and Accumulation Functions

#6.4 The Fundamental Theorem of Calculus and Accumulation Functions

Now that we have introduced integrals, you may be wondering how they connect with derivatives. Today, we’ll introduce the Fundamental Theorem of Calculus which connects these two topics.

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#∫ Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration. It states that the derivative of an integral is just the function inside the integral. 😊

We can define a new function, $F$, that represents the antiderivative of $f$.

$$F(x)=\int_{a}^{x}f(t)dt$$

The Fundamental Theorem of Calculus states that if $f$ is continuous on the interval $(a,b)$ then for every $x$ in the interval:

$$\frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=F'(x)=f(x)$$

This means that we can use the Fundamental Theorem of Calculus to find derivatives. Let’s walk through an example to see what this means.

#✏️ Fundamental Theorem of Calculus Example

Find $g'(16)$ if $g(x)$ is the function below.

$$g(x)=\int_{5}^{x}\sqrt[4]{t}dt$$

How should we proceed? If we ignore what $g(x)$ is specifically defined as for a minute and think about how we would proceed for a function that is not defined with integrals, we would do so by first finding the derivative of $g(x)$ and then plugging $16$ in. Well for functions defined by definite integrals, also called accumulation functions, we do the same!

How do we differentiate $\int_{5}^{x}\sqrt[4]{t}dt$?

This is where the Fundamental Theorem of Calculus comes in. Since it states that the derivative of an integral is just the function inside the integral…

$$g'(x)=\sqrt[4]{x}$$

Now all we have to do is substitute 16. $$g'(16)=\sqrt[4]{16}=2$$

Not too bad right? 🪄

#✏️ Fundamental Theorem of Calculus Example 2

In the above example, the upper bound function was just $x$, making it super easy! All we had to do was substitute $x$ into the integrated and multiply by the derivative of $x$.

Questions become slightly more complicated when the functional bound is something other than x, such as the following function. Let’s find $F'(x)$.

$$F(x)=\int_{3}^{x^2}(t+4)dt$$

The upper bound in this question is $x^2$, so we have to multiply our answer by the derivative of the functional bound.

$$F'(x)=\frac{d}{dx}\int_{3}^{x^2}(t+4)dt$$

Let’s use the chain rule.

$$F'(x)=(x^2+4)*\frac{d}{dx}x^2$$

$$F'(x)=(x^2+4)(2x)$$

So, when the upper bound of the definite integral is a function of x, you need to multiply the integral's integrand by the derivative of the upper bound. It’s not too difficult, but it does mean you have to keep an eye on the upper bound! 👀

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#📝 Practice Problems

Time to try some practice on our own!

#❓ Problems

#Question 1

Let $g(x)=\int_{0}^{x}\sqrt{8+\cos(t)}dt$. Find $g'(0)$.

#Question 2

Let $g(x)=\int_{1}^{x}(5t^2+2t)dt$. Find $g'(3)$.

#Question 3

Let $g(x)=\int_{0}^{x}\sqrt{\sin\left(t\right)+15}dt$. Find $g'(\frac{\pi}{2})$.

#Question 4

Let $F(x)=\int_{3x}^{1}sec^2(t)dt$. Find $F'(x)$.

#✅ Answers and Solutions

#Question 1

To find $g'(0)$, we simply have to find $g'(x)$ and evaluate it at $x=0$ since the upper bound is already $x$!

By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{8+\cos(x)}$.

Therefore, $g'(0)=\sqrt{8+\cos(0)}=\sqrt{8+1}=\sqrt{9}=3$.

#Question 2

To find $g'(3)$, we simply have to find $g'(x)$ and evaluate it at $x=3$.

By the Fundamental Theorem of Calculus, $g'(x)=5x^2+2x$.

Therefore, $g'(3)=5\cdot3^2+2\cdot 3=45+6=51$.

#Question 3

To find $g'(\frac{\pi}{2})$, we simply have to find $g'(x)$ and evaluate it at $x=\frac{\pi}{2}$.

By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{\sin(x)+15}$.

Therefore, $g'(\frac{\pi}{2})=\sqrt{15+\sin(\frac{\pi}{2})}=\sqrt{15+1}=\sqrt{16}=4$.

#Question 4

To find $F'(x)$, we first have to notice that the upper bound is not x.

The upper bound needs to have the variable, so we have to use integral rules to switch the bounds. If you’re going through our guides in order, you haven’t come across integral rules yet. We’re covering that in two key topics, 6.6! Here’s the integral rule you’d be using in this question:

$$\int_{a}^{b}f(x), dx = -\int_{b}^{a}f(x), dx$$

Here’s how we apply it…

$$F'(x)=\frac{d}{dx}(-\int_{1}^{3x}sec^2(t)dt)$$

And here’s the rest of the problem, using what we already know!

$$F'(x)=-sec^2(3x)*\frac{d}{dx}3x$$

$$=-3sec^2(3x)$$

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#💫 Closing

Great work! You've learned about the Fundamental Theorem of Calculus, which, remember, bridges the gap between differentiation and integration.

You've also seen how to apply this theorem to find derivatives of functions defined by definite integrals, including cases where the upper bound is a function of $x$. When the upper bound is not a simple $x$, you multiply the integrand by the derivative of the upper bound.

Keep practicing and working through the practice problems to solidify your understanding. Calculus can be challenging, but with dedication and practice, you'll master it! Good luck. 📚