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?
 A: 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 $$\begin{align}
A=&\int_{-12}^{5}60-7x-x^2dx\\
=&60x-\frac{7x^2}{2}-\frac{x^3}{3}\mid_{-12}^{5}\\=&\frac{4913}{6}.
\end{align}$$
A: 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 $$\int_{-12}^5 (60 - 7x - x^2) \, dx$$
A: Here's how to find the area between two functions, $f(x)$ and $g(x)$.


*

*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).

*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:
$$ \int_a^b f(x) = F(b) - F(a) \text{ where } F'(x) = f(x) $$
$$ 1020 - 7\color{green}{\left(\frac{5^2}{2} - \frac{(-12)^2}{2} \right)} - \color{blue}{\left(\frac{5^3}{3} - \frac{(-12)^3}{3}\right)} $$
$$ 1020 - 7\color{green}{(25/2 - 144/2)} - \color{blue}{(125/3 + 1728/3)} $$
$$ = 9683 / 3$$
