What is the value of the integral from 0 to 1 of x^2 dx?

What is the value of the integral from 0 to 1 of x^2 dx?

• A. 1/2
• B. 1/3
• C. 1/4
• D. 1/5

Answer: (B)

To find the value of the integral from 0 to 1 of \(x^2\) dx, we first need to find the antiderivative of the function \(x^2\). The general formula for integrating \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration. In this case, \(n=2\). Calculating the antiderivative: \[ \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} + C \] Now, we will evaluate the definite integral from 0 to 1: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 \] Next, we substitute the limits into the antiderivative: \[ = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3} \] Thus, the value of the integral from

To find the value of the integral from 0 to 1 of (x^2) dx, we first need to find the antiderivative of the function (x^2). The general formula for integrating (x^n) is (\frac{x^{n+1}}{n+1} + C), where (C) is the constant of integration. In this case, (n=2).

Calculating the antiderivative:

[

\int x^2 , dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} + C

]

Now, we will evaluate the definite integral from 0 to 1:

[

\int_0^1 x^2 , dx = \left[ \frac{x^3}{3} \right]_0^1

]

Next, we substitute the limits into the antiderivative:

[

= \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}

]

Thus, the value of the integral from