How do you evaluate the definite integral $\int (2 x^{3}) d x$ $int (2x^3)dx$ from [1,3]?

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How do you evaluate the definite integral $\int (2 x^{3}) d x$ $int (2x^3)dx$ from [1,3]?

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Answer 1 · 2016-12-11T10:03:38

$\int_{1}^{3} 2 x^{3} d x = 40$ $int_1^3 2x^3dx = 40$

Explanation:

First, we will consider the integral as I with the limits:

$I = \int_{1}^{3} 2 x^{3} d x$ $I = int_1^3 2x^3dx$

We will take out the constant as it is in multiplication with the variable:

$I = 2 \int_{1}^{3} x^{3} d x$ $I = 2 int_1^3 x^3dx$

We know the power rule of integration:

$I = \int x^{n} d x = \frac{x^{n + 1}}{n + 1}$ $I = int x^ndx = (x^(n+1))/(n+1)$

Applying power rule on the integral:

$I = 2 {[\frac{x^{3 + 1}}{3 + 1}]}_{1}^{3}$ $I = 2 [(x^(3+1))/(3+1)]_1^3$

$I = 2 {[\frac{x^{4}}{4}]}_{1}^{3}$ $I = 2 [(x^4)/4]_1^3$

Now we can put the limits of the integration, we know the rule of limits of integration:

Upper Limit-Lower Limit

Hence:

$I = 2 [(\frac{3^{4}}{4}) - (\frac{1^{4}}{4})]$ $I = 2 [((3^4)/4)-(1^4/4)]$

We know, $3^{4} = 81$ $3^4=81$ and $1^{4} = 1$ $1^4=1$

$I = 2 [(\frac{81}{4}) - (\frac{1}{4})]$ $I = 2 [((81)/4)-(1/4)]$

As the base is same, we can directly subtract $81$ $81$ and $1$ $1$

$I = 2 [(\frac{81 - 1}{4})]$ $I = 2 [((81-1)/4)]$

$I = 2 [\frac{80}{4}]$ $I = 2 [80/4]$

We can divide $80$ $80$ by $4$ $4$

$I = 2.20$ $I = 2.20$

$I = 40$ $I=40$

Hence:

$\int_{1}^{3} 2 x^{3} d x = 40$ $int_1^3 2x^3dx = 40$

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