I have been trying to solve a homework problem that reads: "The region between the graphs of $f(x) = x^{2}+2$ and $g(x) = -5x+2$ has what area?" I have been trying to solve this problem and have come up with the answer of $185/6$ square units, but it has come to my attention that that is not correct. So, I was hoping maybe someone can help me out with this problem. Thank you so much!
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If you show your work, we may be able to help. I get $125/6$ – Ross Millikan Feb 18 '14 at 20:48
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There is probably an arithmetical error somewhere. The area is $\int_{-5}^0 -(5x+x^2),dx$. I think this is $\frac{125}{6}$ – André Nicolas Feb 18 '14 at 20:52
3 Answers
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If you compute the intersection of the two graphs, you get the value as $x=-5$ and $x=0$.
Then, I am assuming you have learnt double integration. The area will simply be the integration of $1$ over the region. \begin{align} &=\int_{x=-5}^{x=0}\int_{y=-5x+2}^{y=x^2+2} 1\cdot dy \cdot dx \\ &=\int_{x=-5}^{x=0} (x^2 + 5x) \cdot dx \\ &= \frac{-125}{6}\\ \therefore \mbox{Area} &= \frac{125}{6} \end{align} (I think there is a error in your question as Andre points out in the comments)
Nitish
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$f$ and $g$ intersect at $x=-5$ and $x=0$, so you have to integrate
$$\int _{-5}^0 dx (g(x)-f(x))=\int _{-5}^0 dx (-5x-x^2)=\left(-5\frac{x^2}{2}-\frac{x^3}{3}\right)_{-5} ^0$$
$$=5^3(\tfrac{1}{2}-\tfrac{1}{3})=\frac{5^3}{6}=\frac{125}{6}$$
flonk
- 2,434
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The two graphs intersect at the points where their $x$-coordinates are equal and their $y$-coordinates are equal. Their $y$-coordinates are $x^2+2$ and $-5x+2$. These are equal when $$ x^2+2=-5x+2. $$ Hence $x^2=-5x$. If $x\ne 0$ then we can divide both sides by $x$ and get $x=-5$. If $x=0$, then that is also a solution, as seen by substitution: $0^2 \overset{\text{?}}{=} -5\cdot 0$.
When $x$ is between $-5$ and $0$, then $x^2+2$ is smaller than $-5x+2$, as seen by plugging in any number between $-5$ and $0$. So $$ \begin{align} & \phantom{{}=} \int_{-5}^0 (\text{bigger function minus smaller function}) \\[8pt] & = \int_{-5}^0 (-5x+2)-(x^2+2) \,dx = \int_{-5}^0 -5x -x^2 \,dx \\[8pt] & = \left.\frac{-5x^2}{2} -\frac{x^3}{3}\right|_{x=-5}^{x=0} = \frac{125}{2} -\frac{125}{3} = \frac{125}{6}. \end{align} $$
