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Sketch the region between $y=x^2+x−2$ and the $x$-axis over the interval $[−4,2]$. Find the area of the region.

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    $\begingroup$ Ok. What have you tried? Where do you get stuck? $\endgroup$
    – The Count
    Jan 21, 2017 at 22:20

3 Answers 3

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You just calculate the zeroes of the function: $0=x^2+x-2$ gives $x_0=-2;x_1=1$.

Then you split the area up in 3 parts and calculate the sum of their absolute values:

$$Area = |\int_{-4}^{-2} \! x^2+x-2 \, \mathrm{d}x| + |\int_{-2}^1 \! x^2+x-2 \, \mathrm{d}x| + |\int_1^2 \! x^2+x-2 \, \mathrm{d}x|$$

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Hint

$$x^2+x-2=(x-1)(x+2)$$

So,

$$\text{Area}=\int_{-4}^{-2}(x-1)(x+2)\,dx-\int_{-2}^{1}(x-1)(x+2)\,dx+\int_{1}^{2}(x-1)(x+2)\,dx=$$

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Well, the area should be

$$\begin{align}Area&=\int_{-4}^2\left|x^2+x-2\right|\ dx\\&=\int_{-4}^{-2}x^2+x-2\ dx-\int_{-2}^1x^2+x-2\ dx+\int_1^2x^2+x-2\ dx\\&=\left.\frac13x^3+\frac12x^2-2x\right]_{-4}^{-2}-\left(\left.\frac13x^3+\frac12x^2-2x\right)\right]_{-2}^2+\left.\frac13x^3+\frac12x^2-2x\right]_1^2\\&=15\end{align}$$

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  • $\begingroup$ That's wrong. You are treating the area below the $x$-axis as negative; but the question simply asks for the area of the region. $\endgroup$
    – TonyK
    Jan 21, 2017 at 22:21
  • $\begingroup$ @TonyK Yes, I'm sorry, think I fixed it. $\endgroup$ Jan 21, 2017 at 22:21
  • $\begingroup$ Your diagram is missing the interval $(1,2]$. So you're not quite there. $\endgroup$
    – TonyK
    Jan 21, 2017 at 22:24
  • $\begingroup$ @TonyK Ok, I think I've fixed it all up 100% XD Ah, I am stupid... $\endgroup$ Jan 21, 2017 at 22:32
  • $\begingroup$ This is a good example of spoon feeding an OP who asked a very poor question (remember, we've been talking about "how to answer", and I believe you mentioned something about a good answer answering a good question?) Okay, so you gave the OP a break. And subsequently: Take a look at the OPs newer question. $\endgroup$
    – amWhy
    Feb 4, 2017 at 22:50

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