The Fundamental Theorem of Calculus

Home → The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) shows that differentiation and integration are inverse processes.

Part 1 (FTC1)

If $f$ is a continuous function on $[a, b],$ then the function $g$ defined by

g (x) = \int_{a}^{x} f (t) d t, a \leq x \leq b

is an antiderivative of $f$ , that is

g^{'} (x) = f (x) or \frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x) .

If $f$ happens to be a positive function, then $g (x)$ can be interpreted as the area under the graph of $f$ from $a$ to $x .$

The first part of the theorem says that if we first integrate $f$ and then differentiate the result, we get back to the original function $f .$

Part 2 (FTC2)

The second part of the fundamental theorem tells us how we can calculate a definite integral.

If $f$ is a continuous function on $[a, b]$ and $F$ is an antiderivative of $f,$ that is $F^{'} = f,$ then

\int_{a}^{b} f (x) d x = F (b) - F (a) or \int_{a}^{b} F^{'} (x) d x = F (b) - F (a) .

To evaluate the definite integral of a function $f$ from $a$ to $b,$ we just need to find its antiderivative $F$ and compute the difference between the values of the antiderivative at $b$ and $a .$

So the second part of the fundamental theorem says that if we take a function $F,$ first differentiate it, and then integrate the result, we arrive back at the original function, but in the form $F (b) - F (a) .$

Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes.

The Area under a Curve and between Two Curves

The area under the graph of the function $f (x)$ between the vertical lines $x = a,$ $x = b$ (Figure $2$ ) is given by the formula

S = \int_{a}^{b} f (x) d x = F (b) - F (a) .

Let $F (x)$ and $G (x)$ be antiderivatives of functions $f (x)$ and $g (x),$ respectively.

If $f (x) \geq g (x)$ on the closed interval $[a, b],$ then the area between the curves $y = f (x),$ $y = g (x)$ and the lines $x = a,$ $x = b$ (Figure $3$ ) is given by

S = \int_{a}^{b} [f (x) - g (x)] d x = F (b) - G (b) - F (a) + G (a) .

The Method of Substitution for Definite Integrals

The definite integral $\int_{a}^{b} f (x) d x$ of the variable $x$ can be changed into an integral with respect to $t$ by making the substitution $x = g (t) :$

\int_{a}^{b} f (x) d x = \int_{c}^{d} f (g (t)) g^{'} (t) d t .

The new limits of integration for the variable $t$ are given by the formulas

c = g^{- 1} (a), d = g^{- 1} (b),

where $g^{- 1}$ is the inverse function to $g,$ that is $t = g^{- 1} (x) .$

Integration by Parts for Definite Integrals

In this case the formula for integration by parts looks as follows:

\int_{a}^{b} u d v = {u v |}_{a}^{b} - \int_{a}^{b} v d u,

where ${u v |}_{a}^{b}$ means the difference between the product of functions $u v$ at $x = b$ and $x = a .$

Solved Problems

Click or tap a problem to see the solution.

Example 1

Calculate the derivative of the function $g (x) = \int_{1}^{x} \sqrt{t^{3} + 4 t} d t$ at $x = 2.$

Example 2

Calculate the derivative of the function $g (x) = \int_{- \frac{π}{2}}^{x} \sqrt{\sin^{2} t + 2} d t$ at $x = \frac{π}{6} .$

Example 3

Find the derivative of the function $g (x) = \int_{3}^{x^{2}} \frac{d t}{t} .$

Example 4

Find the derivative of the function $g (x) = \int_{0}^{x^{2}} \sqrt{1 + t^{2}} d t .$

Example 5

Find the derivative of the function $g (x) = \int_{1}^{x^{3}} t^{2} d t .$

Example 6

Find the derivative of the function $g (x) = \int_{x^{2}}^{x^{3}} t d t .$

Example 7

Calculate the derivative of the function $g (x) = \int_{\sqrt{x}}^{x} (t^{2} - t) d t$ at $x = 1.$

Example 8

Evaluate the integral $\int_{0}^{2} (x^{3} - x^{2}) d x .$

Example 1.

Calculate the derivative of the function $g (x) = \int_{1}^{x} \sqrt{t^{3} + 4 t} d t$ at $x = 2.$

Solution.

We apply the Fundamental Theorem of Calculus, Part $1 :$

g^{'} (x) = \frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x) .

Hence

g^{'} (x) = \frac{d}{d x} (\int_{1}^{x} \sqrt{t^{3} + 4 t} d t) = \sqrt{x^{3} + 4 x} .

Substituting $x = 2$ yields

g^{'} (2) = \sqrt{2^{3} + 4 \cdot 2} = \sqrt{16} = 4.

Example 2.

Calculate the derivative of the function $g (x) = \int_{- \frac{π}{2}}^{x} \sqrt{\sin^{2} t + 2} d t$ at $x = \frac{π}{6} .$

Solution.

We use the Fundamental Theorem of Calculus, Part $1 :$

g^{'} (x) = \frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x) .

Then

g^{'} (x) = \frac{d}{d x} (\int_{- \frac{π}{2}}^{x} \sqrt{\sin^{2} t + 2} d t) = \sqrt{\sin^{2} x + 2} .

Note that the lower limit of integration $- \frac{π}{2}$ does not affect the answer.

Now we compute the value of the derivative for $x = \frac{π}{6} :$

g^{'} (\frac{π}{6}) = \sqrt{\sin^{2} \frac{π}{6} + 2} = \sqrt{{(\frac{1}{2})}^{2} + 2} = \sqrt{\frac{9}{4}} = \frac{3}{2} .

Example 3.

Find the derivative of the function $g (x) = \int_{3}^{x^{2}} \frac{d t}{t} .$

Solution.

We introduce the new function

h (u) = \int_{3}^{u} \frac{d t}{t} .

Using the FTC1, we have

h^{'} (u) = \frac{1}{u} .

As $g (x) = h (x^{2}),$ then by the chain rule

g^{'} (x) = {[h (x^{2})]}^{'} = h^{'} (x^{2}) \cdot {(x^{2})}^{'} = h^{'} (x^{2}) \cdot 2 x = \frac{1}{x^{2}} \cdot 2 x = \frac{2}{x} .

Example 4.

Find the derivative of the function $g (x) = \int_{0}^{x^{2}} \sqrt{1 + t^{2}} d t .$

Solution.

Since the upper limit of integration is not $x,$ we apply the chain rule. Let $u = x^{2},$ then $u^{'} = 2 x .$

Consider the new function

h (u) = \int_{0}^{u} \sqrt{1 + t^{2}} d t .

By the FTC1, we can write

h^{'} (u) = \sqrt{1 + u^{2}} .

As $g (x) = h (x^{2}),$ we have

g^{'} (x) = {[h (x^{2})]}^{'} = h^{'} (x^{2}) \cdot {(x^{2})}^{'} = \sqrt{1 + {(x^{2})}^{2}} \cdot 2 x = 2 x \sqrt{1 + x^{4}} .

Example 5.

Find the derivative of the function $g (x) = \int_{1}^{x^{3}} t^{2} d t .$

Solution.

Let $u = x^{3},$ then $u^{'} = 3 x^{2} .$

We introduce the new function

h (u) = \int_{0}^{u} t^{2} d t .

Using the FTC1, we obtain

h^{'} (u) = u^{2} .

Since $g (x) = h (x^{3}),$ we have

g^{'} (x) = {[h (x^{3})]}^{'} = h^{'} (x^{3}) \cdot {(x^{3})}^{'} = {(x^{3})}^{2} \cdot 3 x^{2} = x^{6} \cdot 3 x^{2} = 3 x^{8} .

Example 6.

Find the derivative of the function $g (x) = \int_{x^{2}}^{x^{3}} t d t .$

Solution.

We split the interval of integration $[x^{2}, x^{3}]$ using an intermediate point $c,$ so that $c \in [x^{2}, x^{3}] .$ Hence the derivative of $g (x)$ is written in the form

g^{'} (x) = \frac{d}{d x} (\int_{x^{2}}^{x^{3}} t d t) = \frac{d}{d x} (\int_{x^{2}}^{c} t d t + \int_{c}^{x^{3}} t d t) = \frac{d}{d x} (\int_{c}^{x^{3}} t d t - \int_{c}^{x^{2}} t d t) = \frac{d}{d x} (\int_{c}^{x^{3}} t d t) - \frac{d}{d x} (\int_{c}^{x^{2}} t d t) .

We calculate both terms using the FTC1 and the chain rule:

\frac{d}{d x} (\int_{c}^{x^{3}} t d t) = x^{3} \cdot {(x^{3})}^{'} = x^{3} \cdot 3 x^{2} = 3 x^{5};

\frac{d}{d x} (\int_{c}^{x^{2}} t d t) = x^{2} \cdot {(x^{2})}^{'} = x^{2} \cdot 2 x = 2 x^{3} .

Then

g^{'} (x) = 3 x^{5} - 2 x^{3} .

Example 7.

Calculate the derivative of the function $g (x) = \int_{\sqrt{x}}^{x} (t^{2} - t) d t$ at $x = 1.$

Solution.

We split the integral function into two terms:

g (x) = \int_{\sqrt{x}}^{x} (t^{2} - t) d t = \int_{\sqrt{x}}^{c} (t^{2} - t) d t + \int_{c}^{x} (t^{2} - t) d t = \int_{c}^{x} (t^{2} - t) d t - \int_{c}^{\sqrt{x}} (t^{2} - t) d t,

where $c \in [x^{2}, x^{3}] .$

Find the derivative of $g (x)$ using the FTC1 and the chain rule (for the second term):

\frac{d}{d x} \int_{c}^{x} (t^{2} - t) d t = x^{2} - x;

\frac{d}{d x} \int_{c}^{\sqrt{x}} (t^{2} - t) d t = ({(\sqrt{x})}^{2} - \sqrt{x}) \cdot {(\sqrt{x})}^{'} = (x - \sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} = \frac{\sqrt{x}}{2} - \frac{1}{2} .

Then

g^{'} (x) = (x^{2} - x) - (\frac{\sqrt{x}}{2} - \frac{1}{2}) = x^{2} - x - \frac{\sqrt{x}}{2} + \frac{1}{2} .

At the point $x = 1,$ the derivative is equal to

g^{'} (1) = 1^{2} - 1 - \frac{\sqrt{1}}{2} + \frac{1}{2} = 0.

Example 8.

Evaluate the integral $\int_{0}^{2} (x^{3} - x^{2}) d x .$

Solution.

Using the Fundamental Theorem of Calculus, Part $2,$ we have

\int_{0}^{2} (x^{3} - x^{2}) d x = {(\frac{x^{4}}{4} - \frac{x^{3}}{3}) |}_{0}^{2} = (\frac{16}{4} - \frac{8}{3}) - 0 = \frac{4}{3} .