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Find the area bounded by

$$ f(x) = x + 6 $$ $$ g(x) = x^3 $$ $$ h(x) = -\frac{x}{2} $$

Edit Fixed simple error on h(x)

I already drew the grew, although it is very hard to really tell where they seem to intersect.

It looks like on the left side of the graph f(x) intersects with h(x) at -12. However going toward the right side of the graph it looks like I'll have to integrate base on f(x) and g(x).

I'm thinking this is a two part integration problem, but I'm unsure of what my bounds should be.

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  • $\begingroup$ Can you see the region bounded below and above by two graphs and bounded on left and right by two values? $\endgroup$ – Sigur Oct 26 '13 at 18:27
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    $\begingroup$ Are you sure that the functions are right? $\endgroup$ – Sigur Oct 26 '13 at 18:28
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    $\begingroup$ $y=-\frac{x}{2}$ $\endgroup$ – J. W. Perry Oct 26 '13 at 18:32
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    $\begingroup$ So, $y=-x/2$. !! $\endgroup$ – Sigur Oct 26 '13 at 18:32
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    $\begingroup$ Now it's clear with $y=-x/2$ instead of $y=x/2$. $\endgroup$ – Julien Oct 26 '13 at 18:34
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Here is the graphs to help you. Decompose your region in two parts: one for $-4\leq x\leq 0$ and the other for $0\leq x\leq 2$. Then you just use integrals to compute the area.

First interval (left): below red minus below green

Second interval (right): below red minus below blue

enter image description here

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  • $\begingroup$ So I can take $\int_{-4}^{0} f(x) - h(x) $ + $\int_0^2 f(x) - g(x)$? $\endgroup$ – ConfusingCalc Oct 26 '13 at 18:39
  • $\begingroup$ From zero to two, not eight. $\endgroup$ – Sigur Oct 26 '13 at 18:40
  • $\begingroup$ Sorry, it was a typo, but I fixed it :) Thanks I'll give this a try! Should I be using delta y for this? $\endgroup$ – ConfusingCalc Oct 26 '13 at 18:42
  • $\begingroup$ @ConfusingCalc, what do you mean by using delta y? Are you asking about the integration variable? you have to integrate with respect to $x$, that is, $dx$. $\endgroup$ – Sigur Oct 26 '13 at 18:44
  • $\begingroup$ Your assumption is correct. I was asking because I get confused when I should be integrating in terms of x or y when dealing with area and volume. $\endgroup$ – ConfusingCalc Oct 26 '13 at 18:46
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Given the image of the graph that @julien posted, we see that the two lines intersect in the second quadrant at $(-4, 2)$. The cubic equation intersects with the line $y = x+6$ in the first quadrant at $(2, 8)$. And the point at which the bounding lower curve switches from $y = -\dfrac x2$ to $y = x^3$ is at $x = 0$, and specifically, at the origin. The bounding upper curve remains the same throughout: $y = x+ 6$.

$$\int_{-4}^0 \left((x + 6) + \frac x2\right)\,dx +\int_0^2 ((x + 6) - x^3)\,dx$$

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  • $\begingroup$ Thanks for explaining amWhy :) the graph on here was a much better presentation than my hand drawn one! $\endgroup$ – ConfusingCalc Oct 26 '13 at 18:50
  • $\begingroup$ You're welcome, @ConfusingCalc! $\endgroup$ – Namaste Oct 26 '13 at 18:51

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