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
\[S = \int\limits_a^b {f\left( x \right)dx} = {F\left( b \right) – F\left( a \right).}\]

Area under a curve

Figure 1.

Area between two curves

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