Review your knowledge of the fundamental theorem of calculus and use it to solve problems.

What is the fundamental theorem of calculus?

The theorem has two versions.

a) ddxaxf(t)dt=f(x)\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)

We start with a continuous function ff and we define a new function for the area under the curve y=f(t)y=f(t):
F(x)=axf(t)dtF(x)=\displaystyle\int_a^x f(t)\,dt
What this version of the theorem says is that the derivative of FF is ff. In other words, FF is an antiderivative of ff. Thus, the theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

b) abf(x)dx=F(b)F(a)\displaystyle\int_a^b\!\! f(x)dx=F(b)\!-\!\!F(a)

This version gives more direct instructions to finding the area under the curve y=f(x)y=f(x) between x=ax=a and x=bx=b. Simply find an antiderivative FF and take F(b)F(a)F(b)-F(a).
Want to learn more about the fundamental theorem of calculus? Check out this video.

Practice set 1: Applying the theorem

Problem 1.1
g(x)=1x2t+7dt\displaystyle g(x) = \int_{\,1}^{\,x}\sqrt{2t+7}\,dt\,
g(9)=\displaystyle g\,^\prime(9)\, =
Want to try more problems like this? Check out this exercise.

Practice set 2: Applying the theorem with chain rule

We can use the theorem in more hairy situations. Let's find, for example, the expression for ddx0x3sin(t)dt\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt. Note that the interval is between 00 and x3x^3, not xx.
To help us, we define F(x)=0xsin(t)dt\displaystyle F(x) = \int_{0}^{x}\sin(t) \, dt. According to the fundamental theorem of calculus, F(x)=sin(x)F'(x)=\sin(x).
It follows from our definition that 0x3sin(t)dt\displaystyle\int_{0}^{x^3}\sin(t) \, dt is F(x3)F(x^3), which means that ddx0x3sin(t)dt\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt is ddxF(x3)\dfrac{d}{dx}F(x^3). Now we can use the chain rule:
=ddx0x3sin(t)dt=ddxF(x3)=F(x3)ddx(x3)=sin(x3)3x2\begin{aligned} &\phantom{=}\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt \\\\ &=\dfrac{d}{dx}F(x^3) \\\\ &=F'(x^3)\cdot\dfrac{d}{dx}(x^3) \\\\ &=\sin(x^3)\cdot 3x^2 \end{aligned}
Problem 2.1
F(x)=0x4cos(t)dt\displaystyle F(x) = \int_{0}^{x^4}\cos(t) \, dt
F(x)=F'(x) =
Want to try more problems like this? Check out this exercise.
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