# How do you evaluate #int [x(x^2+1)]dx# for [0, 1]?

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To evaluate the integral ( \int_0^1 x(x^2 + 1) , dx ), you can use the properties of definite integrals and basic integration techniques. First, expand the expression inside the integral. Then, integrate the expanded expression term by term. Finally, evaluate the result at the upper and lower limits of integration and subtract the value at the lower limit from the value at the upper limit.

Expanding the expression ( x(x^2 + 1) ) gives ( x^3 + x ). Integrating term by term, the antiderivative of ( x^3 ) is ( \frac{1}{4}x^4 ) and the antiderivative of ( x ) is ( \frac{1}{2}x^2 ).

So, ( \int_0^1 x(x^2 + 1) , dx = \left[\frac{1}{4}x^4 + \frac{1}{2}x^2 \right]_0^1 ).

Evaluating the antiderivative at the upper and lower limits:

At ( x = 1 ), the value is ( \frac{1}{4}(1)^4 + \frac{1}{2}(1)^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} ).

At ( x = 0 ), the value is ( \frac{1}{4}(0)^4 + \frac{1}{2}(0)^2 = 0 ).

Subtracting the value at the lower limit from the value at the upper limit:

( \frac{3}{4} - 0 = \frac{3}{4} ).

Therefore, ( \int_0^1 x(x^2 + 1) , dx = \frac{3}{4} ).

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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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