Area between two curves

The area between two graphs is the integral of the upper function minus the lower one.

Check which is on top on each interval, and use the $x$-coordinates of the intersections as the limits. If the curves swap places, split the interval.

Examples

1. 01For 0<x<3, which expression gives the area between these two curves?$\displaystyle y=3x,\quad y=x^2$AnswerHide

   Answer$\displaystyle \int_{0}^{3} (3x-x^2)\,dx$

   Steps

   1. $\displaystyle \int_{0}^{3} (3x-x^2)\,dx$$\displaystyle = \frac{9}{2}$
2. 02The area enclosed by $y=9$ and $y=x^2$. What goes in □?$\displaystyle \int_{-3}^{3} (\square)\,dx$AnswerHide

   Answer$\displaystyle 9-x^2$

   Steps

   1. $\displaystyle 9 - x^2 = 0$$\displaystyle \Rightarrow x = \pm 3$
   2. The crossings become the limits
   3. $\displaystyle \int_{-3}^{3} (9-x^2)\,dx$$\displaystyle = 36$
3. 03Which expression gives the area between these two curves?$\displaystyle y=4,\quad y=x^2$AnswerHide

   Answer$\displaystyle \int_{-2}^{2} (4-x^2)\,dx$

   Steps

   1. $\displaystyle 4 - x^2 = 0$$\displaystyle \Rightarrow x = \pm 2$
   2. The crossings become the limits
   3. $\displaystyle \int_{-2}^{2} (4-x^2)\,dx$$\displaystyle = \frac{32}{3}$
4. 04The curves meet at x=0 and x=4, and y=4x is on top between them. What is the enclosed area?$\displaystyle y=4x,\quad y=x^2$AnswerHide

   Answer$\displaystyle \frac{32}{3}$
5. 05The area enclosed by $y=1$ and $y=x^2$. What goes in □?$\displaystyle \int_{-1}^{1} (\square)\,dx$AnswerHide

   Answer$\displaystyle 1-x^2$

   Steps

   1. $\displaystyle 1 - x^2 = 0$$\displaystyle \Rightarrow x = \pm 1$
   2. The crossings become the limits
   3. $\displaystyle \int_{-1}^{1} (1-x^2)\,dx$$\displaystyle = \frac{4}{3}$
6. 06Which expression gives the area between these two curves?$\displaystyle y=16,\quad y=x^2$AnswerHide

   Answer$\displaystyle \int_{-4}^{4} (16-x^2)\,dx$

   Steps

   1. $\displaystyle 16 - x^2 = 0$$\displaystyle \Rightarrow x = \pm 4$
   2. The crossings become the limits
   3. $\displaystyle \int_{-4}^{4} (16-x^2)\,dx$$\displaystyle = \frac{256}{3}$

FAQ