Question

How do you integrate ${\left( {2x - 5} \right)^2}dx?$

Answer 1

Hint: To integrate the given expression, first of all expand the given expression either with use binomial expansion or with use of the algebraic identity for the square of difference between two numbers. Then use the distributive property of integration to differentiate each term of the expansion separately and then simplify them up to get the required integration.
Algebraic identity for square of difference between two numbers is given as
${(a - b)^2} = {a^2} - 2ab + {b^2}$

Complete step-by-step solution:
In order to integrate the given expression ${\left( {2x - 5} \right)^2}dx$ we will first expand the expression ${\left( {2x - 5} \right)^2}$ as following
We will use the algebraic identity for square of difference between two numbers in order to expand the given expression, which is given as
$\Rightarrow {(a - b)^2} = {a^2} - 2ab + {b^2}$
On comparing this identity with the expression, we get
\Rightarrow $a = 2x\;{\text{and}}\;b = 5$
So expanding the expression as follows
$
  \Rightarrow {(2x - 5)^2} = {(2x)^2} - 2 \times 2x \times 5 + {5^2} \\
    \Rightarrow 4{x^2} - 20x + 25 \\
 $
Now we can write the given integration as
$\Rightarrow \int {{{\left( {2x - 5} \right)}^2}dx} = \int {\left( {4{x^2} - 20x + 25} \right)dx} $
Using distributive property of integration, in order to distributive over each terms of the expansion, we will get
$\Rightarrow \int {\left( {4{x^2} - 20x + 25} \right)dx} = \int {4{x^2}dx} - \int {20dx} + \int {25dx} $
Integrating each term in order to solve further, we will get
$
 \Rightarrow \int {4{x^2}dx} - \int {20xdx} + \int {25dx} = \dfrac{{4{x^3}}}{3} - \dfrac{{20{x^2}}}{2} + 25x + c \\
  \Rightarrow \dfrac{{4{x^3}}}{3} - 10{x^2} + 25x + c \\
 $

Therefore $\dfrac{{4{x^3}}}{3} - 10{x^2} + 25x + c$ is the required integration of the expression ${\left( {2x - 5} \right)^2}$


Note: After integration of the indefinite integrals always write $ \pm c$ in its integration part, this $ \pm c$ is an arbitrary constant or the constant of integration. Let us see the necessity of writing $ \pm c$ at the integration part with an example, derivative of ${x^2} + 23$ is $2x$ and the derivative of ${x^2} - 2$ is also $2x$, so in order to maintain the constant in the integration part, we write $ \pm c$.