Question

What is the value of the integral of x^3 from 0 to 2?

Answer 1

To find the value of the integral of
`{x}^(3)` from 0 to 2, we need to evaluate the definite integral:

`\int_(0)^(2) {x}^(3) dx`

Using the power rule of integration, we can integrate the function as follows:

`\int {x}^(3) dx = \frac{{x}^(4)}{4} + C`

where C is the constant of integration.

Plugging in the limits of integration, we get:

`\int_(0)^(2) {x}^(3) dx = \left[\frac{{x}^(4)}{4}\right]_(0)^(2)`

`= \left(\frac{{2}^(4)}{4}\right) - \left(\frac{{0}^(4)}{4}\right)`

`= (16)/(4) - 0`

`= 4`

Therefore, the value of the integral of
`{x}^(3)` from 0 to 2 is 4.

Answer 2

Answer: 4

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Work Shown:

`\displaystyle \int_(0)^(2)\text{x}^3d\text{x}\\\\\\\displaystyle (1)/(4)\text{x}^(4)+C\bigg|_(0)^(2)\\\\\\\displaystyle \left[(1)/(4)2^(4)+C\right]-\left[(1)/(4)0^(4)+C\right]\\\\\\4\\\\\\\text{Therefore,} \ \displaystyle \int_(0)^(2)\text{x}^3d\text{x} = 4\\\\`

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The integral rule I used in the 2nd step was

`\displaystyle \int \text{x}^(n)d\text{x} = (1)/(n+1)\text{x}^(n+1)+C\\\\`

in which you can verify it's valid by differentiating
`(1)/(n+1)\text{x}^(n+1)+C\\\\` to end up with
`\text{x}^(n)` again.