-3
$\begingroup$

enter image description here

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

$\endgroup$

closed as off-topic by Namaste, Noah Schweber, Simply Beautiful Art, pjs36, C. Falcon Feb 5 '17 at 0:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Noah Schweber, Simply Beautiful Art, pjs36, C. Falcon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Ok. What have you tried? Where do you get stuck? $\endgroup$ – The Count Jan 21 '17 at 22:20
0
$\begingroup$

Well, the area should be

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

enter image description here

$\endgroup$
  • $\begingroup$ 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 '17 at 22:21
  • $\begingroup$ @TonyK Yes, I'm sorry, think I fixed it. $\endgroup$ – Simply Beautiful Art Jan 21 '17 at 22:21
  • $\begingroup$ Your diagram is missing the interval $(1,2]$. So you're not quite there. $\endgroup$ – TonyK Jan 21 '17 at 22:24
  • $\begingroup$ @TonyK Ok, I think I've fixed it all up 100% XD Ah, I am stupid... $\endgroup$ – Simply Beautiful Art Jan 21 '17 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. $\endgroup$ – Namaste Feb 4 '17 at 22:50
1
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

Hint

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

So,

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.