Mathematics

# Evaluate the following integral:$\int { \left( 2-3x \right) \left( 3+2x \right) \left( 1-2x \right) } dx$

##### SOLUTION
We have $\left(2-3x\right)\left(3+2x\right)\left(1-2x\right)$

$=\left(2\left(3+2x\right)-3x\left(3+2x\right)\right)\left(1-2x\right)$

$=\left(6+4x-9x-6{x}^{2}\right)\left(1-2x\right)$

$=\left(6-5x-6{x}^{2}\right)\left(1-2x\right)$

$=6\left(1-2x\right)-5x\left(1-2x\right)-6{x}^{2}\left(1-2x\right)$

$=6-12x-5x+10{x}^{2}-6{x}^{2}+12{x}^{3}$

$=12{x}^{3}+4{x}^{2}-17x+6$

Now $\displaystyle\int{\left(2-3x\right)\left(3+2x\right)\left(1-2x\right)dx}$

$=\displaystyle\int{\left(12{x}^{3}+4{x}^{2}-17x+6\right)dx}$

We know that $\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$

$=12\dfrac{{x}^{4}}{4}+4\dfrac{{x}^{3}}{3}-17\dfrac{{x}^{2}}{2}+6x+c$ where $c$ is the constant of integration

$=3{x}^{4}+\dfrac{4}{3}{x}^{3}-\dfrac{17}{2}{x}^{2}+6x+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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