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# Calculus

Integration of Functions

# The Definite Integral and Fundamental Theorem of Calculus

Page 1
Problems 1-3
Page 2
Problems 4-8

Let $$f\left( x \right)$$ be a continuous function on the closed interval $$\left[ {a,b} \right].$$ Divide this interval into $$n$$ partial subintervals. Choose a sample point $${\xi_i}$$ in each subinterval and make up an integral sum $$\sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}}$$, where $$\Delta {x_i}$$ is the length of the $$i$$th subinterval.

The definite integral of $$f\left( x \right)$$ from $$a$$ to $$b$$ is defined to be the limit
$\require{AMSmath.js} {\int\limits_a^b {f\left( x \right)dx} }={ \lim\limits_{\substack{ n \to \infty \\ \text{max}\,\Delta {x_i} \to 0}} \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} ,}$ where $$\Delta {x_i} = {x_i} – {x_{i – 1}},$$ $${x_{i – 1}} \le {\xi _i} \le {x_i}.$$

### Properties of the Definite Integral

We assume below that $$f\left( x \right)$$ and $$g\left( x \right)$$ are continuous functions on the closed interval $$\left[ {a,b} \right].$$

1. $$\int\limits_a^b {1dx} = b – a$$
2. $$\int\limits_a^b {kf\left( x \right)dx}$$ $$= k \int\limits_a^b {f\left( x \right)dx} ,$$ where $$k$$ is a constant;
3. $$\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx}$$ $$= \int\limits_a^b {f\left( x \right)dx}$$ $$+{ \int\limits_a^b {g\left( x \right)dx} }$$
4. $$\int\limits_a^b {f\left( x \right)dx}$$ $$= \int\limits_a^c {f\left( x \right)dx}$$ $$+ \int\limits_c^b {f\left( x \right)dx} ,$$ where $$a \lt c \lt b;$$
5. If $$0 \le f\left( x \right) \le g\left( x \right)$$ for all $$x \in \left[ {a,b} \right],$$ then $$0 \le \int\limits_a^b {f\left( x \right)dx}$$ $$\le \int\limits_a^b {g\left( x \right)dx} .$$
6. $$\int\limits_a^a {f\left( x \right)dx} = 0$$
7. $$\int\limits_a^b {f\left( x \right)dx}$$ $$= – \int\limits_b^a {f\left( x \right)dx}$$
8. If $$f\left( x \right) \ge 0$$ in the interval $$\left[ {a,b} \right],$$ then $$\int\limits_a^b {f\left( x \right)dx} \ge 0$$

### The Fundamental Theorem of Calculus

Let $$f\left( x \right)$$ be a function, which is continuous on the closed interval $$\left[ {a,b} \right].$$ If $$F\left( x \right)$$ is any antiderivative of $$f\left( x \right)$$ on $$\left[ {a,b} \right],$$ then
${\int\limits_a^b {f\left( x \right)dx} } = {\left. {F\left( x \right)} \right|_a^b } = {F\left( b \right) – F\left( a \right).}$

### 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 $$1$$) is given by the formula
$S = \int\limits_a^b {f\left( x \right)dx} = {F\left( b \right) – F\left( a \right).}$

Figure 1.

Figure 2.

Let $$F\left( x \right)$$ and $$G\left( x \right)$$ be indefinite integrals 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 $$2$$) 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
${c = {g^{ – 1}}\left( a \right),\;\;}\kern-0.3pt{d = {g^{ – 1}}\left( b \right),}$ where $${g^{ – 1}}$$ is the inverse function to $$g,$$ i.e. $$t = {g^{ – 1}}\left( x \right).$$

### 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 on problem description to see solution.

### ✓Example 1

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

### ✓Example 2

Calculate the integral $$\int\limits_0^1 {\left( {\sqrt[\large 3\normalsize]{t} – \sqrt t } \right)dt} .$$

### ✓Example 3

Evaluate the integral $$\int\limits_0^1 {{\large\frac{x}{{{{\left( {3{x^2} – 1} \right)}^4}}}\normalsize} dx}.$$

### ✓Example 4

Evaluate the integral $$\int\limits_0^{\ln 2} {x{e^{ – x}}dx}.$$

### ✓Example 5

Find the area bounded by the curves $$y = {x^2}$$ and $$y = \sqrt x .$$

### ✓Example 6

Find the area bounded by the curves $$y = 2x – {x^2}$$ and $$x + y = 0.$$

### ✓Example 7

Find the area of the triangle with vertices at $$\left( {0,0} \right),$$ $$\left( {2,6} \right)$$ and $$\left( {7,1} \right).$$

### ✓Example 8

Find the area inside the ellipse $${\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} = 1.$$

### Example 1.

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

#### Solution.

Using the fundamental theorem of calculus, we have
${\int\limits_0^2 {\left( {{x^3} – {x^2}} \right)dx} } = {\left. {\left( {\frac{{{x^4}}}{4} – \frac{{{x^3}}}{3}} \right)} \right|_0^2 } = {\left( {\frac{{16}}{4} – \frac{8}{3}} \right) – 0 }={ \frac{4}{3}.}$

### Example 2.

Calculate the integral $$\int\limits_0^1 {\left( {\sqrt[\large 3\normalsize]{t} – \sqrt t } \right)dt} .$$

#### Solution.

${\int\limits_0^1 {\left( {\sqrt[\large 3\normalsize]{t} – \sqrt t } \right)dt} } = {\int\limits_0^1 {\left( {{t^{\large\frac{1}{3}\normalsize}} – {t^{\large\frac{1}{2}\normalsize}}} \right)dt} } = {\left. {\left( {\frac{{{t^{\large\frac{1}{3}\normalsize + 1}}}}{{\frac{1}{3} + 1}} – \frac{{{t^{\large\frac{1}{2}\normalsize + 1}}}}{{\frac{1}{2} + 1}}} \right)} \right|_0^1 } = {\left. {\left( {\frac{{3{t^{\large\frac{4}{3}\normalsize}}}}{4} – \frac{{2{t^{\large\frac{3}{2}\normalsize}}}}{3}} \right)} \right|_0^1 } = {\left( {\frac{3}{4} – \frac{2}{3}} \right) – 0 }={ \frac{1}{{12}}.}$

### Example 3.

Evaluate the integral $$\int\limits_0^1 {{\large\frac{x}{{{{\left( {3{x^2} – 1} \right)}^4}}}\normalsize} dx}.$$

#### Solution.

First we make the substitution:
${t = 3{x^2} – 1,\;\;}\Rightarrow {dt = 6xdx,\;\;}\Rightarrow {xdx = \frac{{dt}}{6}.}$ Determine the new limits of integration. When $$x = 0,$$ then $$t = -1.$$ When $$x = 1,$$ then we have $$t = 2.$$ So, the integral with the new variable $$t$$ can be easily calculated:
${\int\limits_0^1 {\frac{x}{{{{\left( {3{x^2} – 1} \right)}^4}}}dx} } = {\int\limits_{ – 1}^2 {\frac{{\frac{{dt}}{6}}}{{{t^4}}}} } = {\frac{1}{6}\int {{t^{ – 4}}dt} } = {\frac{1}{6}\left. {\left( {\frac{{{t^{ – 3}}}}{{ – 3}}} \right)} \right|_{ – 1}^2 } = { – \frac{1}{{18}}\left( {\frac{1}{8} – 1} \right) } = {\frac{7}{{144}}.}$

Page 1
Problems 1-3
Page 2
Problems 4-8