Question

integration of (3x²-4)x dx​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\int\:(\:3x^3\:-\:4x\:)\:dx\:=\:\dfrac{x^2\:(\:3x^2\:-\:8\:)}{4}\:+\:C}}}$

Step-by-step-explanation:

We have to integrate ( 3x² - 4 ) x.

( 3x² - 4 ) x

⇒ 3x³ - 4x

Now, integrating with respect to x, we get,

$\displaystyle{\sf\:\int\:(\:3x^3\:-\:4x\:)\:dx}$

$\displaystyle{\implies\sf\:\int\:3x^3\:dx\:-\:\int\:4x\:dx}$

We know that,

$\displaystyle{\boxed{\sf\:\int\:k\:f\:(x)\:=\:k\:\int\:f\:(x)}}$

$\displaystyle{\implies\sf\:3\:\int\:x^3\:-\:4\:\int\:x}$

We know that,

$\displaystyle{\boxed{\sf\:\int\:x^n\:=\:\dfrac{x^{n\:+\:1}}{n\:+\:1}}}$

$\displaystyle{\implies\sf\:3\:\dfrac{x^{3\:+\:1}}{3\:+\:1}\:-\:4\:\dfrac{x^{1\:+\:1}}{1\:+\:1}}$

$\displaystyle{\implies\sf\:3\:\dfrac{x^4}{4}\:-\:4\:\dfrac{x^2}{2}}$

$\displaystyle{\implies\sf\:\dfrac{3\:x^4}{4}\:-\:\dfrac{4\:x^2}{2}\:\times\:\dfrac{2}{2}}$

$\displaystyle{\implies\sf\:\dfrac{3\:x^4}{4}\:-\:\dfrac{8\:x^2}{4}}$

$\displaystyle{\implies\sf\:\dfrac{3\:x^4\:-\:8\:x^2}{4}}$

$\displaystyle{\implies\sf\:\dfrac{x^2\:(\:3x^2\:-\:8\:)}{4}\:+\:C}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\int\:(\:3x^3\:-\:4x\:)\:dx\:=\:\dfrac{x^2\:(\:3x^2\:-\:8\:)}{4}\:+\:C}}}}$