Integrate x^3-6x from 0 to 3

\int_{0}^{3}\left(x^3-6x\right)dx

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Answer

$-\frac{27}{4}$

Step by step solution

Problem

$\int_{0}^{3}\left(x^3-6x\right)dx$
1

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

$\int_{0}^{3}-6xdx+\int_{0}^{3} x^3dx$
2

Taking the constant out of the integral

$\int_{0}^{3} x^3dx-6\int_{0}^{3} xdx$
3

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

$\left[\frac{x^{4}}{4}\right]_{0}^{3}-6\int_{0}^{3} xdx$
4

Evaluate the definite integral

$-6\int_{0}^{3} xdx+\frac{0^{4}}{4}\left(-1\right)+\frac{3^{4}}{4}$
5

Calculate the power

$-6\int_{0}^{3} xdx+\frac{0}{4}\left(-1\right)+\frac{81}{4}$
6

Divide $81$ by $4$

$-6\int_{0}^{3} xdx+0\left(-1\right)+\frac{81}{4}$
7

Any expression multiplied by $0$ is equal to $0$

$-6\int_{0}^{3} xdx+0+\frac{81}{4}$
8

Add the values $\frac{81}{4}$ and $0$

$\frac{81}{4}-6\int_{0}^{3} xdx$
9

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

$\frac{81}{4}-6\left[\frac{1}{2}x^2\right]_{0}^{3}$
10

Evaluate the definite integral

$\left(3^2\cdot 0.5-1\cdot 0^2\cdot 0.5\right)\left(-6\right)+20.25$
11

Multiply $\frac{1}{2}$ times $-1$

$\left(0^2\left(-0.5\right)+3^2\cdot 0.5\right)\left(-6\right)+20.25$
12

Calculate the power

$\left(0\left(-0.5\right)+9\cdot 0.5\right)\left(-6\right)+20.25$
13

Any expression multiplied by $0$ is equal to $0$

$\left(0+9\cdot 0.5\right)\left(-6\right)+20.25$
14

Multiply $\frac{1}{2}$ times $9$

$\left(0+4.5\right)\left(-6\right)+20.25$
15

Add the values $\frac{9}{2}$ and $0$

$4.5\left(-6\right)+20.25$
16

Multiply $-6$ times $\frac{9}{2}$

$20.25-27$
17

Subtract the values $\frac{81}{4}$ and $-27$

$-\frac{27}{4}$

Answer

$-\frac{27}{4}$

Problem Analysis

Main topic:

Integral calculus

Time to solve it:

0.22 seconds

Views:

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