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Integrate 3x^2-5 from 3 to 6

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Answer

$174$

Step by step solution

Problem

$\int_{3}^{6}\left(3x^2-5\right)dx$
1

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

$\int_{3}^{6}-5dx+\int_{3}^{6}3x^2dx$
2

The integral of a constant is equal to the constant times the integral's variable

$\left[-5x\right]_{3}^{6}+\int_{3}^{6}3x^2dx$
3

Evaluate the definite integral

$\int_{3}^{6}3x^2dx-1\cdot 3\left(-5\right)+6\left(-5\right)$
4

Multiply $-5$ times $6$

$\int_{3}^{6}3x^2dx+15-30$
5

Subtract the values $15$ and $-30$

$\int_{3}^{6}3x^2dx-15$
6

Taking the constant out of the integral

$3\int_{3}^{6} x^2dx-15$
7

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

$\left[3\frac{x^{3}}{3}\right]_{3}^{6}-15$
8

Simplify the fraction

$\left[x^{3}\right]_{3}^{6}-15$
9

Evaluate the definite integral

$-15+3^{3}\left(-1\right)+6^{3}$
10

Calculate the power

$-15+27\left(-1\right)+216$
11

Subtract the values $216$ and $-15$

$27\left(-1\right)+201$
12

Multiply $-1$ times $27$

$201-27$
13

Subtract the values $201$ and $-27$

$174$

Answer

$174$

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Main topic:

Integral calculus

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