# How do you evaluate the definite integral #int (x+1)dx# from [1,5]?

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To evaluate the definite integral (\int_{1}^{5} (x+1) , dx), you can use the fundamental theorem of calculus. First, integrate the function ((x+1)) with respect to (x) to get ( \frac{x^2}{2} + x ). Then, evaluate this antiderivative at the upper bound (5) and subtract the value at the lower bound (1).

[ \begin{align*} \int_{1}^{5} (x+1) , dx &= \left[\frac{x^2}{2} + x\right]_{1}^{5} \ &= \left(\frac{5^2}{2} + 5\right) - \left(\frac{1^2}{2} + 1\right) \ &= \left(\frac{25}{2} + 5\right) - \left(\frac{1}{2} + 1\right) \ &= \left(\frac{25}{2} + \frac{10}{2}\right) - \left(\frac{1}{2} + \frac{2}{2}\right) \ &= \left(\frac{35}{2}\right) - \left(\frac{3}{2}\right) \ &= \frac{35}{2} - \frac{3}{2} \ &= \frac{32}{2} \ &= 16 \end{align*} ]

So, the value of the definite integral is (16).

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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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