# The Fundamental Theorem of Calculus

## Trigonometry # 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 $$\left[ {a,b} \right],$$ then the function $$g$$ defined by

$g\left( x \right) = \int\limits_a^x {f\left( {t} \right)dt},\;\; a \le x \le b$

is an antiderivative of $$f$$, that is

$g^\prime\left( x \right) = f\left( x \right)\;\;\text{or}\;\;\frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) = f\left( x \right).$

If $$f$$ happens to be a positive function, then $$g\left( x \right)$$ 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 $$\left[ {a,b} \right]$$ and $$F$$ is an antiderivative of $$f,$$ that is $$F^\prime = f,$$ then

$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)\;\;\text{or}\;\;\int\limits_a^b {{F^\prime\left( x \right)}dx} = F\left( b \right) - F\left( a \right).$

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 $$F\left( b \right) - F\left( a \right).$$

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 $$f\left( x \right)$$ between the vertical lines $$x = a,$$ $$x = b$$ (Figure $$2$$) is given by the formula

$S = \int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right).$

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

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

$S = \int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = F\left( b \right) - G\left( b \right) - F\left( a \right) + G\left( a \right).$

## The Method of Substitution for Definite Integrals

The definite integral $$\int\limits_a^b {f\left( x \right)dx}$$ of the variable $$x$$ can be changed into an integral with respect to $$t$$ by making the substitution $$x = g\left( t \right):$$

$\int\limits_a^b {f\left( x \right)dx} = \int\limits_c^d {f\left( {g\left( t \right)} \right)g'\left( t \right)dt} .$

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

$c = {g^{ - 1}}\left( a \right),\;\;d = {g^{ - 1}}\left( b \right),$

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

## Integration by Parts for Definite Integrals

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

$\int\limits_a^b {udv} = \left. {uv} \right|_a^b - \int\limits_a^b {vdu} ,$

where $$\left. {uv} \right|_a^b$$ means the difference between the product of functions $$uv$$ 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 $g\left( x \right) = \int\limits_1^x {\sqrt {{t^3} + 4t} dt}$ at $$x = 2.$$

### Example 2

Calculate the derivative of the function $g\left( x \right) = \int\limits_{ - \frac{\pi }{2}}^x {\sqrt {{{\sin }^2}t + 2} dt}$ at $$x = \frac{\pi }{6}.$$

### Example 3

Find the derivative of the function $g\left( x \right) = \int\limits_3^{{x^2}} {\frac{{dt}}{t}}.$

### Example 4

Find the derivative of the function $g\left( x \right) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}.$

### Example 5

Find the derivative of the function $g\left( x \right) = \int\limits_1^{{x^3}} {{t^2}dt}.$

### Example 6

Find the derivative of the function $g\left( x \right) = \int\limits_{{x^2}}^{{x^3}} {tdt}.$

### Example 7

Calculate the derivative of the function $g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} - t} \right)dt}$ at $$x = 1.$$

### Example 8

Evaluate the integral $\int\limits_0^2 {\left( {{x^3} - {x^2}} \right)dx}.$

### Example 1.

Calculate the derivative of the function $g\left( x \right) = \int\limits_1^x {\sqrt {{t^3} + 4t} dt}$ at $$x = 2.$$

Solution.

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

$g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) = f\left( x \right).$

Hence

$g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_1^x {\sqrt {{t^3} + 4t} dt} } \right) = \sqrt {{x^3} + 4x} .$

Substituting $$x = 2$$ yields

$g^\prime\left( 2 \right) = \sqrt {{2^3} + 4 \cdot 2} = \sqrt {16} = 4.$

### Example 2.

Calculate the derivative of the function $g\left( x \right) = \int\limits_{ - \frac{\pi }{2}}^x {\sqrt {{{\sin }^2}t + 2} dt}$ at $$x = \frac{\pi }{6}.$$

Solution.

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

$g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) = f\left( x \right).$

Then

$g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_{ - \frac{\pi }{2}}^x {\sqrt {{{\sin }^2}t + 2} dt} } \right) = \sqrt {{{\sin }^2}x + 2} .$

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

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

$g^\prime\left( {\frac{\pi }{6}} \right) = \sqrt {{{\sin }^2}\frac{\pi }{6} + 2} = \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + 2} = \sqrt {\frac{9}{4}} = \frac{3}{2}.$

### Example 3.

Find the derivative of the function $g\left( x \right) = \int\limits_3^{{x^2}} {\frac{{dt}}{t}}.$

Solution.

We introduce the new function

$h\left( u \right) = \int\limits_3^u {\frac{{dt}}{t}}.$

Using the FTC1, we have

$h^\prime\left( u \right) = \frac{1}{u}.$

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

$g^\prime\left( x \right) = \left[ {h\left( {{x^2}} \right)} \right]^\prime = h^\prime\left( {{x^2}} \right) \cdot \left( {{x^2}} \right)^\prime = h^\prime\left( {{x^2}} \right) \cdot 2x = \frac{1}{{{x^2}}} \cdot 2x = \frac{2}{x}.$

### Example 4.

Find the derivative of the function $g\left( x \right) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}.$

Solution.

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

Consider the new function

$h\left( u \right) = \int\limits_0^u {\sqrt {1 + {t^2}} dt} .$

By the FTC1, we can write

$h^\prime\left( u \right) = \sqrt {1 + {u^2}} .$

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

$g^\prime\left( x \right) = \left[ {h\left( {{x^2}} \right)} \right]^\prime = h^\prime\left( {{x^2}} \right) \cdot \left( {{x^2}} \right)^\prime = \sqrt {1 + {{\left( {{x^2}} \right)}^2}} \cdot 2x = 2x\sqrt {1 + {x^4}} .$

### Example 5.

Find the derivative of the function $g\left( x \right) = \int\limits_1^{{x^3}} {{t^2}dt}.$

Solution.

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

We introduce the new function

$h\left( u \right) = \int\limits_0^u {{t^2}dt} .$

Using the FTC1, we obtain

$h^\prime\left( u \right) = {u^2}.$

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

$g^\prime\left( x \right) = \left[ {h\left( {{x^3}} \right)} \right]^\prime = h^\prime\left( {{x^3}} \right) \cdot \left( {{x^3}} \right)^\prime = {\left( {{x^3}} \right)^2} \cdot 3{x^2} = {x^6} \cdot 3{x^2} = 3{x^8}.$

### Example 6.

Find the derivative of the function $g\left( x \right) = \int\limits_{{x^2}}^{{x^3}} {tdt}.$

Solution.

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

$g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_{{x^2}}^{{x^3}} {tdt} } \right) = \frac{d}{{dx}}\left( {\int\limits_{{x^2}}^c {tdt} + \int\limits_c^{{x^3}} {tdt} } \right) = \frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} - \int\limits_c^{{x^2}} {tdt} } \right) = \frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} } \right) - \frac{d}{{dx}}\left( {\int\limits_c^{{x^2}} {tdt} } \right).$

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

$\frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} } \right) = {x^3} \cdot \left( {{x^3}} \right)^\prime = {x^3} \cdot 3{x^2} = 3{x^5};$
$\frac{d}{{dx}}\left( {\int\limits_c^{{x^2}} {tdt} } \right) = {x^2} \cdot \left( {{x^2}} \right)^\prime = {x^2} \cdot 2x = 2{x^3}.$

Then

$g^\prime\left( x \right) = 3{x^5} - 2{x^3}.$

### Example 7.

Calculate the derivative of the function $g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} - t} \right)dt}$ at $$x = 1.$$

Solution.

We split the integral function into two terms:

$g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} - t} \right)dt} = \int\limits_{\sqrt x }^c {\left( {{t^2} - t} \right)dt} + \int\limits_c^x {\left( {{t^2} - t} \right)dt} = \int\limits_c^x {\left( {{t^2} - t} \right)dt} - \int\limits_c^{\sqrt x } {\left( {{t^2} - t} \right)dt},$

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

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

$\frac{d}{{dx}}\int\limits_c^x {\left( {{t^2} - t} \right)dt} = {x^2} - x;$
$\frac{d}{{dx}}\int\limits_c^{\sqrt x } {\left( {{t^2} - t} \right)dt} = \left( {{{\left( {\sqrt x } \right)}^2} - \sqrt x } \right) \cdot \left( {\sqrt x } \right)^\prime = \left( {x - \sqrt x } \right) \cdot \frac{1}{{2\sqrt x }} = \frac{{\sqrt x }}{2} - \frac{1}{2}.$

Then

$g^\prime\left( x \right) = \left( {{x^2} - x} \right) - \left( {\frac{{\sqrt x }}{2} - \frac{1}{2}} \right) = {x^2} - x - \frac{{\sqrt x }}{2} + \frac{1}{2}.$

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

$g^\prime\left( 1 \right) = {1^2} - 1 - \frac{{\sqrt 1 }}{2} + \frac{1}{2} = 0.$

### Example 8.

Evaluate the integral $\int\limits_0^2 {\left( {{x^3} - {x^2}} \right)dx}.$

Solution.

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

$\int\limits_0^2 {\left( {{x^3} - {x^2}} \right)dx} = \left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^3}}}{3}} \right)} \right|_0^2 = \left( {\frac{{16}}{4} - \frac{8}{3}} \right) - 0 = \frac{4}{3}.$