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

Answer 1

Christopher Agresta

#1/2#

#int_1^2(x^2-1)/(x+1)dx=int_1^2((x+1)(x-1))/(x+1)dx# =#int_1^2(x-1)dx=x^2/2-x|_1^2# =#2^2/2-2-(1/2-1)# =#1/2#

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Answer 2

Samantha Adkison

To evaluate the integral (\int_{1}^{2} \frac{x^2 - 1}{x + 1} , dx) over the interval [1, 2], you can first rewrite the integrand as (\frac{x^2 - 1}{x + 1} = x - 1). Then, integrate (x - 1) with respect to (x) over the interval [1, 2].

(\int_{1}^{2} (x - 1) , dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2})

Now substitute the upper and lower limits of integration:

(\left[ \frac{(2)^2}{2} - 2 \right] - \left[ \frac{(1)^2}{2} - 1 \right] = \left[ \frac{4}{2} - 2 \right] - \left[ \frac{1}{2} - 1 \right])

(= (2 - 2) - (\frac{1}{2} - 1) = 0 - (\frac{1}{2} - 1))

(= -\frac{1}{2} + 1 = \frac{1}{2})

Therefore, the value of the integral (\int_{1}^{2} \frac{x^2 - 1}{x + 1} , dx) over the interval [1, 2] is (\frac{1}{2}).

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Answer from HIX Tutor

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.