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I have a little Calculus problem which is confusing me quite a bit, so I thought to ask you guys for help. The problem consists in calculating the area between three curves, they are:

$$ - y = x² - 6x + 8 $$ $$- y = 2x - 4 $$ $$ - y = x + 2 $$

Here's the graph of those three functions plotted: problem_graph

I'd like to know what's the value of the area, what are the integrals that reach that value and most importantly, since I cannot see it properly, which is the area that should be calculated between these three curves? (Below the red curve and above the blue curve? Below the yellow curve and above the blue curve? Below the yellow curve and above the red curve?)

Thanks for you attention!

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    $\begingroup$ I would interpret it as the region between the two lines and inside the blue curve. $\endgroup$ – Ted Shifrin Oct 11 '16 at 4:35
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    $\begingroup$ I agree with Ted, as that region uses all three curves as a border at some part of the region. $\endgroup$ – turkeyhundt Oct 11 '16 at 4:36
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I think it obvious that it can only be the area that is bounded by all three curves. If we denote

$$ y_1=8-6x+x^2\\ y_2=-4+2x\\ y_3=2+x $$

then we can express the area as

$$A=\int_1^2 (y_2-y_1)~dx + \int_2^6 (y_2-y_3)~dx$$

I'll assume you can handle it from here. FYI, I solved this as shown as well numerically by a completely different method and found that the area is $A=10\frac{1}{6}$.

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graphed functions

$y_1=8-6x+x^2$ (blue curve) bounds the side, $y_2=-4+2x$ (red curve) is the lower bound, and $y_3=2+x$ (yellow curve) is the upper bound.

This gives us three areas:

  1. Between Red and Blue
  2. Between Yellow and Blue
  3. Between Yellow and Red, bounded to the left by Blue

To find Area 1, use the following formula: $$ \int_a^b [f_1(x) - f_2(x)]dx $$ Where $f_1(x)$ is the upper bound ($y_3$ or yellow) and $f_2(x)$ is the lower bound ($y_1$ or blue).

$a$ and $b$, respectively, are the leftmost and rightmost intersections of $f_1(x)$ and $f_2(x)$.

Solve: $$ \int_1^6 [(2 + x) - ( 8 - 6x + x^2)] dx\\ =\int_1^6 (7x-6-x^2) dx\\ =[\frac{7x^2}{2} - 6x - \frac{x^3}{3}]_1^6\\ =(\frac{7(6)^2}{2} - 6(6) - \frac{(6)^3}{3}) - (\frac{7(1)^2}{2} - 6(1) - \frac{(1)^3}{3})\\ =(126 - 36 - 72) - (\frac{7}{2} - 6 - \frac{1}{3}) = 18 + \frac{17}{6}\\ =20\frac{5}{6} $$

To find Area 2, use the same method you used for area one, but using $y_2$ as your upper bound, or $f_1(x)$.

Solve: $$ \int_2^6 [(-4+2x) - ( 8 - 6x + x^2)]dx\\ =\int_2^6 (8x-12- x^2)dx\\ =[4x^2 - 12x - \frac{x^3}{3}]_2^6\\ =(4(6)^2 - 12(6) - \frac{(6)^3}{3}) - (4(2)^2 - 12(2) - \frac{(2)^3}{3})\\ =(144 - 72 - 72) - (16 - 24 - \frac{8}{3})=0 - (- 10\frac{2}{3})\\ =10\frac{2}{3} $$

Area 3 is Area 2 sans Area 1, so subtract the two areas.

$20\frac{5}{6} - 10\frac{2}{3} = 10\frac{1}{6}$

$10\frac{1}{6}$ is your answer!

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