Integral of (4x-1x^2)x

\int\left(4x-x^2\right) xdx\pi^2

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Answer

$-\sqrt{6}x^{4}+13.1595x^{3}+C_0$

Step by step solution

Problem

$\int\left(4x-x^2\right) xdx\pi^2$
1

Calculate the power

$9.8696\int x\left(4x-x^2\right)dx$
2

Multiply $\left(4x+-x^2\right)$ by $x$

$9.8696\int\left(4x^2-x^{3}\right)dx$
3

The integral of a sum of two or more functions is equal to the sum of their integrals

$9.8696\left(\int-x^{3}dx+\int4x^2dx\right)$
4

Taking the constant out of the integral

$9.8696\left(\int4x^2dx-\int x^{3}dx\right)$
5

Taking the constant out of the integral

$9.8696\left(4\int x^2dx-\int x^{3}dx\right)$
6

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$9.8696\left(4\int x^2dx-\frac{x^{4}}{4}\right)$
7

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$9.8696\left(4\frac{x^{3}}{3}-\frac{x^{4}}{4}\right)$
8

Simplify the fraction

$9.8696\left(\frac{4}{3}x^{3}-\frac{x^{4}}{4}\right)$
9

Multiply $\left(\frac{4}{3}x^{3}+-\frac{x^{4}}{4}\right)$ by $9.8696$

$13.1595x^{3}-9.8696\frac{x^{4}}{4}$
10

Simplify the fraction

$13.1595x^{3}-\sqrt{6}x^{4}$
11

Add the constant of integration

$-\sqrt{6}x^{4}+13.1595x^{3}+C_0$

Answer

$-\sqrt{6}x^{4}+13.1595x^{3}+C_0$

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Problem Analysis

Main topic:

Integral calculus

Time to solve it:

0.97 seconds

Views:

90