# Second Fundamental Theorem of Calculus Hello and welcome to this video about the Second Fundamental Theorem of Calculus! We will explore what this part says and how it works, but first a refresher of the First Part:

If f(x) is continuous on [a,b] and $$\int f(x)dx=F(x)$$ on [a,b], then

$$\int_a^bf(x)dx=F(b)-F(a)$$

Let’s get right to it! The Second Part of the Fundamental Theorem of Calculus (It’s got ‘Fundamental’ in its name, so you know it’s important – in fact, it’s perhaps the most important in calculus!) is closely related to the first and says:

If f(x) is is continuous on [a,b] and the function F(x) is defined by

$$F(x)=\int_a^xf(t)dt$$

then F'(x)=f(x) over [a,b].

The biggest implication with this part is that it guarantees that every integrable function has an antiderivative. Notice how F(x) is defined as an integral function – x can be any value and the function will return a number. This is what’s called an accumulation function.

Let’s see how this part of the theorem works!

Suppose $$g(x)=\int_1^xf(t) dt$$. Use the given values of F(t) to evaluate g(x).

F(0)=0 F(1) =2 F(2)=-3 F(4)=5

 x g(x) 0 $$\int_1^0f(t) dt=F(0)-F(1)=-2$$ 1 $$\int_1^1F(t) dt=0$$ 2 $$\int_1^2f(t) dt=F(2)-F(1)=-5$$ 4 $$\int_1^4f(t) dt=F(4)-F(1)=3$$

Let’s try another one!

If $$hx=\int_2^x(3t^2+4t)dt$$, then, by the theorem, we know $$h’x=3x^2+4x$$. Notice the derivative function is expressed in terms of x instead of t, since the integral function is a function of x.

Why is this true? Well, let’s take a look!

$$h(x)=\int_2^x(3t^2+4t)dt$$
=$$t^3+2t^2|_2^x$$
=$$(x^3+2x^2)-(2^3+2(2)^2$$
=$$x^3+2x^2-16$$

Therefore, $$h’x=3x^2+4x$$.

This is especially powerful with more complex antiderivatives like this:

If $$r(x)=\int_3^x\frac{\sqrt{t}}{t^2-3}dt$$, then $$r'(x)=\frac{\sqrt{x}}{x^2-3}dx$$

Let’s get a little practice in.

If $$g(x)=\int_{-5}^{2x}(t^2-t-3)dt$$, find a formula for g(x) and calculate g'(x).

$$g(x)=\frac{1}{3}t^3-\frac{1}{2}t^2-3t|_{-5}^{2x}$$

$$g(x)=(\frac{8x^3}{3}-\frac{4x^2}{2}-6x)-(-\frac{125}{3}-\frac{25}{2}+15)$$
$$g(x)=\frac{8x^3}{3}-\frac{4x^2}{2}-6x+\frac{235}{6}$$
$$g'(x)=8x^2-4x-6$$

Suppose $$g(x)=\int{-2}^{x}f(t) dt$$. Use the provided values of F(t) to evaluate g(x).
F(-1)=6 F(-5)=0 F(2)=10 F(-2)=4

 x g(x) -1 $$\int_{-2}^{-1}f(t) dt=F(-1)-F(-2)=6-4=2$$ -5 $$\int_{-2}^{-5}f(t) dt=F(-5)-F(-2)=0-4=-4$$ 2 $$\int_{-2}^{2}f(t) dt=F(2)-F(-2)=10-4=6$$

I hope that this video helped with your understanding of the Second Fundamental Theorem of Calculus!

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See you next time!