Question

Evaluate the following definite integrals as a limit of sums.
1. $\int_a^b {xdx} $
2. $\int_0^5 {\left( {x + 1} \right)dx} $
3. $\int_2^3 {{x^2}} dx$
4. $\int_1^4 {\left( {{x^2} - x} \right)dx} $
5. $\int_{ - 1}^1 {{e^x}dx} $
6. $\int_0^4 {\left( {x + {e^{2x}}} \right)} dx$

Answer 1

Hint: Integration is a method of finding the value of function g(x) and the derivative of g(x), Dg(x), is equal to another function f(x). Integration is mainly used to find areas between curves, volumes and central points such as centre of mass, centre of gravity etc. Integration can be solved by many methods. Integration by substitution and Integration by parts are the most used ones. We will use integration by parts in this solution.

Complete step-by-step answer:
We are given to solve 6 definite integrals.
A definite integral is an integral with upper and lower limits.
Every integral is given with a lower limit which is written at the bottom of the integration sign and an upper limit which is written at the top of the integration sign.
1. $\int_a^b {xdx} $
Integration of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$, where n can be any integer or a fraction.
$
  \int_a^b {xdx} = \int_a^b {{x^1}dx} \\
  n = 1 \\
   \to \int_a^b {xdx} = \left[ {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right]_a^b \\
  = \left[ {\dfrac{{{x^2}}}{2}} \right]_a^b \\
  = \left[ {\left( {\dfrac{{{b^2}}}{2}} \right) - \left( {\dfrac{{{a^2}}}{2}} \right)} \right] \\
  = \dfrac{{{b^2}}}{2} - \dfrac{{{a^2}}}{2} = \dfrac{{{b^2} - {a^2}}}{2} \\
 $
2. $\int_0^5 {\left( {x + 1} \right)dx} $
This integral can be solved using Integration by parts.
We are just expanding the sum and separating the integral.
$\int_0^5 {\left( {x + 1} \right)dx} = \int_0^5 {xdx} + \int_0^5 {1dx} $
Integration of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$, where n can be any integer or a fraction.
$
 \int_0^5 {\left( {x + 1} \right)dx} = \int_0^5 {xdx} + \int_0^5 {1dx} \\
 \int_0^5 {xdx} = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^5 \\
 = \left[ {\left( {\dfrac{{{5^2}}}{2}} \right) - \left( {\dfrac{{{0^2}}}{2}} \right)} \right] \\
   = \dfrac{{25}}{2} - 0 = \dfrac{{25}}{2} \\
  \int_0^5 {1dx} = \int_0^5 {{x^0}dx} \\
  \int_0^5 {{x^0}dx} = \left[ {\dfrac{{{x^{0 + 1}}}}{{0 + 1}}} \right]_0^5 \\
   = \left[ x \right]_0^5 = 5 - 0 = 5 \\
  \int_0^5 {\left( {x + 1} \right)dx} = \dfrac{{25}}{2} + 5 = \dfrac{{35}}{2} \\
 $
3. $\int_2^3 {{x^2}} dx$
Integration of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$, where n can be any integer or a fraction.
$
  \int_2^3 {{x^2}dx} \\
  n = 2 \\
   \to \int_2^3 {{x^2}dx} = \left[ {\dfrac{{{x^{2 + 1}}}}{{2 + 1}}} \right]_2^3 \\
= \left[ {\dfrac{{{x^3}}}{3}} \right]_2^3 \\
= \left[ {\left( {\dfrac{{{3^3}}}{3}} \right) - \left( {\dfrac{{{2^3}}}{3}} \right)} \right] \\
= \left( {\dfrac{{27}}{3}} \right) - \left( {\dfrac{8}{3}} \right) = \dfrac{{27 - 8}}{3} = \dfrac{{19}}{3} \\
 $
4. $\int_1^4 {\left( {{x^2} - x} \right)dx} $
This integral can be solved using Integration by parts.
We are just expanding the sum and separating the integral.
$\int_1^4 {\left( {{x^2} - x} \right)dx} = \int_1^4 {{x^2}dx} - \int_1^4 {xdx} $
Integration of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$, where n can be any integer or a fraction.
$
  \int_1^4 {\left( {{x^2} - x} \right)dx} = \int_1^4 {{x^2}dx} - \int_1^4 {xdx} \\
  \int_1^4 {{x^2}dx} = \left[ {\dfrac{{{x^3}}}{3}} \right]_1^4 \\
  = \left[ {\left( {\dfrac{{{4^3}}}{3}} \right) - \left( {\dfrac{{{1^3}}}{3}} \right)} \right] \\
  = \dfrac{{64}}{3} - \dfrac{1}{3} = \dfrac{{63}}{3} = 21 \\
  \int_1^4 {xdx} = \int_1^4 {{x^1}dx} \\
  \int_1^4 {{x^1}dx} = \left[ {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right]_1^4 \\
  = \left[ {\dfrac{{{x^2}}}{2}} \right]_1^4 \\
 = \left( {\dfrac{{{4^2}}}{2}} \right) - \left( {\dfrac{{{1^2}}}{2}} \right) = \dfrac{{16}}{2} - \dfrac{1}{2} = \dfrac{{15}}{2} \\
  \int_1^4 {\left( {{x^2} - x} \right)dx} = 21 - \dfrac{{15}}{2} = \dfrac{{27}}{2} \\
 $
5. $\int_{ - 1}^1 {{e^x}dx} $
Integration of an exponent function is the exponent function itself.
$
  \int_{ - 1}^1 {{e^x}dx} = \left[ {{e^x}} \right]_{ - 1}^1 \\
  = {e^1} - {e^{ - 1}} \\
  = e - \dfrac{1}{e} \\
  = \dfrac{{{e^2} - 1}}{e} \\
 $
6. $\int_0^4 {\left( {x + {e^{2x}}} \right)} dx$
This integral can be solved using Integration by parts.
We are just expanding the sum and separating the integral.
$\int_0^4 {\left( {x + {e^{2x}}} \right)} dx = \int_0^4 {xdx} + \int_0^4 {{e^{2x}}dx} $
Integration of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$, where n can be any integer or a fraction.
$
  \int_0^4 {\left( {x + {e^{2x}}} \right)} dx = \int_0^4 {xdx} + \int_0^4 {{e^{2x}}dx} \\
  \int_0^4 {xdx} = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^4 \\
  = \left[ {\left( {\dfrac{{{4^2}}}{2}} \right) - \left( {\dfrac{{{0^2}}}{2}} \right)} \right] \\
  = \dfrac{{16}}{2} - 0 = \dfrac{{16}}{2} = 8 \\
  \int_0^4 {{e^{2x}}dx} = \dfrac{1}{2}\left[ {{e^{2x}}} \right]_0^4 \\
  = \dfrac{1}{2}\left[ {\left( {{e^{2 \times 4}}} \right) - \left( {{e^0}} \right)} \right] \\
  = \dfrac{1}{2}\left( {{e^8} - 1} \right) \\
  \int_0^4 {\left( {x + {e^{2x}}} \right)} dx = 8 + \dfrac{{\left( {{e^8} - 1} \right)}}{2} \\
   = \dfrac{{16 + {e^8} - 1}}{2} = \dfrac{{15 + {e^8}}}{2} \\
 $

Note: Integral is also known as indefinite integral which has no limits or boundaries whereas definite integrals have boundaries. Do not confuse definite integrals with indefinite integrals. Also do not forget to apply limits in definite integrals.