# How do you evaluate the integral #int_0^1x^2dx# ?

Remember one of the most important theorems in Calculus:

Fundamental Theorem of Calculus (Part 2)

Let us the theorem above to evaluate the definite integral.

by plugging in the upper and the lower limits,

by simplifying,

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To evaluate the integral (\int_0^1 x^2 , dx), you can use the fundamental theorem of calculus, which states that the integral of a function over an interval can be found by evaluating its antiderivative at the endpoints of the interval and taking the difference.

- Find the antiderivative of (x^2).
- Evaluate the antiderivative at the upper limit of integration (1).
- Evaluate the antiderivative at the lower limit of integration (0).
- Subtract the value at the lower limit from the value at the upper limit.

[ \int_0^1 x^2 , dx = \left[ \frac{1}{3} x^3 \right]_0^1 = \frac{1}{3}(1^3) - \frac{1}{3}(0^3) = \frac{1}{3} ]

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To evaluate the integral ∫₀¹x² dx, we use the power rule for integration. Integrating x² with respect to x yields (1/3)x³. Evaluating this expression from 0 to 1 gives [(1/3)(1)³] - [(1/3)(0)³]. Simplifying, we get (1/3) - 0, which equals 1/3. Therefore, the value of the integral ∫₀¹x² dx is 1/3.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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