# Area enclosed between the curves $y=x^2$ and $y=60-7x$

Find the area enclosed between the curves $y=x^2$ and $y=60-7x$.

I am completely new at this but I have tried and I believe that it should be between the numbers $0$ and $60$ is this right?

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The area is between 810 and 820. If you show your work someone can help you spot the mistake. –  Joni Mar 17 '13 at 17:05
Calculate intersections of your functions and integrate the function $60-7x-x^2$ with this limits. –  Andre Mar 17 '13 at 17:05

It is always a good idea to draw the functions if possible(i.e if you know the graphs of the functions, and you should learn some basic graphs like lines,sin and cos, roots,etc.), then find the intersection points between the two function which are in this case $x=5,x=-12$ as the other answers show. Here is a graph of these functions

and as the graph shows $60-7x$ is above $x^2$ thus the area is \begin{align} A=&\int_{-12}^{5}60-7x-x^2dx\\ =&60x-\frac{7x^2}{2}-\frac{x^3}{3}\mid_{-12}^{5}\\=&\frac{4913}{6}. \end{align}

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Find where $y = x^2$ and $y = 60 - 7x$ intersect:

You can do this by setting the equations equal to one another and solving the resulting quadratic equation $$x^2 = 60 - 7x \iff x^2 +7x - 60 = 0$$ $$\iff (x+12)(x-5) = 0$$ $$\iff \;x = -12, \;\;\text{or}\;\;x = 5$$

Use these values as your limits of integration, and calculate the integral of the area between the line $y = 60 - 7x$ and the parabola $y = x^2$ by integrating the function $$(60 -7x) -x^2$$

using those $x$-values as limits.

Note: We subtract from $y = 60-7x$, the function $\;y = x^2\;$ because for every $x\in (-12, 5),\;\;60 - 7x > x^2$.

This can easily be shown by graphing the two functions, which is always a good thing to do for problems like this. This helps determine the number of regions over which to integrate, if more than one such region exists, which function is the "upper" and which the "lower", and bounds of integration. WolframAlpha gives us:

That is, to compute the area of the region bound by $\;y = 60 - 7x\;$ and $\;y = x^2,\;$ between the values of intersection $\;x = -12\;$ to $\;x=5,\;$ compute $$\int_{-12}^5 (60 - 7x - x^2) \, dx$$

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Does this make sense? If $,F = \int (60 - 7x - x^2)\,dx\;$ then evaluate $F(5) - F(-12) =$ Area. So compute the integral, then determine its value at $5$, and subtract from that its value at $-12$. –  amWhy Mar 17 '13 at 17:33
Needs positive feedback! +1 –  Amzoti May 4 '13 at 0:40

Here's how to find the area between two functions, $f(x)$ and $g(x)$.

1. Find when $f(x) = g(x)$, calling the intersections $a$ and $b$. Assume $f(x) \ge g(x)$ over the interval $[a, b]$ (this will make sure the area is positive).
2. Find $$\int_{a}^b f(x) - g(x) \; \mathrm dx$$

In your case, we have that:

$$x^2 = 60 - 7x$$ $$x^2 + 7x - 60 = 0$$

The solutions are $-12$ and $5$, and we call these $a$ and $b$. Since $60 - 7x$ is greater over this interval, we integrate:

$$\int_{-12}^{5} 60 - 7x - x^2 \; \mathrm dx$$ $$\int_{-12}^{5} 60 \; \mathrm dx - \int_{-12}^{5} 7x \; \mathrm dx - \int_{-12}^{5} x^2 \; \mathrm dx$$

Consider a rectangle with length $17$ ($|-12| + |5| = 17$) and height $60$ (the integrand). Now the area is $60 \cdot 17 = 1020$.

$$1020 - 7\color{green}{\int_{-12}^{5} x \; \mathrm dx} - \color{blue}{\int_{-12}^{5} x^2 \; \mathrm dx}$$

By the power rule of integration:

$$\int x^n \; \mathrm dx = \frac{x^{n-1}}{n}$$

and the fundamental theorem of calculus:

$$\int_a^b f(x) = F(b) - F(a) \text{ where } F'(x) = f(x)$$

$$1020 - 7\color{green}{\left(\frac{5^2}{2} - \frac{(-12)^2}{2} \right)} - \color{blue}{\left(\frac{5^3}{3} - \frac{(-12)^3}{3}\right)}$$ $$1020 - 7\color{green}{(25/2 - 144/2)} - \color{blue}{(125/3 + 1728/3)}$$ $$= 9683 / 3$$

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Yes this I know but it is the calculation after that I get stuck with –  user1838781 Mar 17 '13 at 17:19
@user1838781, I added more information. Do you know how to integrate functions such as $x^2$ and $x$? What exactly are you having trouble with. –  George V. Williams Mar 17 '13 at 17:24
After the integration I belive that your supposed to calculate so that in the end it's only a simple number –  user1838781 Mar 17 '13 at 17:29
@user1838781, I've completed the problem by using the fundamental theorem of calculus and the power rule for integration. Do you understand how the results are derived now? –  George V. Williams Mar 17 '13 at 17:36
No I don't know how to integrate it. Is it something like: if I take it all together [60-7-2x] what now? –  user1838781 Mar 17 '13 at 17:38