Section 6.1: Area Between Curves
displacement - distance between start and end points
distance traveled - total distance
Example 1:
$disp$ = 5 - 2 = 3
$DT$ = 5 + 2 = 7
$disp \le DT$
Example 2:
$v(t) = 2 - t$, $[0, 3]$
Find $disp$., $DT$
$disp = 2 - \frac{1}{2} = \frac{3}{2}$
$DT = 2 + \frac{1}{2} = \frac{5}{2}$
disp = $\int_0^3 (2 - t) \,dt = 2t - \frac{t^2}{2} \biggr\rvert_{0}^{3}$
$[2(3) - \frac{(3)^2}{2}] - [2(0)-\frac{(0)^2}{2}]$
$6 - \frac{9}{2} = \frac{3}{2}$
$\frac{3}{2}$
Definition
The total area between $f(x)$ and x-axis on $[a, b]$ is $\int_b^a \lvert f(x) \rvert dx$
- Subdivide $[a, b]$ where $f(x)$ changes sign (zeros)
- Find the antiderivative ($F(x)$)
- Add absolute values of integrals on subintervals
Example 3:
$v(t) 3t^3 -18t + 24$, $[0, 5]$
Find displacement and distance traveled
$s(t) = \frac{3t^3}{3} - \frac{18t^2}{1} + 24t = t^3 - 9t^2 + 24t$
$disp = s(t) \biggr\rvert_{0}^{5} = 125 - 225 + 120 = 20$
20
$v(t) = 3(t^2-6t+8) = 3(t-2)(t-4)$
$DT = \int_0^2 \lvert v(t) \rvert dt + \int_2^9 \lvert v(t) \rvert dt + \int_4^5 \lvert v(t) \rvert dt$
$= \int_0^2 v(t) dt$ - $\int_2^4 v(t) dt$ + $\int_4^5 v(t) dt$
$= s(t) \biggr\rvert_{0}^{2} - s(t) \biggr\rvert_{2}^{4} + s(t) \biggr\rvert_{4}^{5}$
$=(s(2)-s(0))-(s(4)-s(2))+(s(5)-s(4))$
$=s(2) + 2s(2) - 2s(4) + s(5)$
$= -0 + 2 * 20 - 2 * 16 + 20 = 28$
28
Example 4:
Find TA for $f(x) = x^\frac{1}{3} - x$ on $[-1, 8]$
$= x^\frac{1}{3}(1-x^\frac{2}{3}) = x^\frac{1}{3}(1+x^\frac{1}{3})(1-x^\frac{1}{3})$
(0) (-1) (1)
$F(x) = \frac{3x^\frac{4}{3}}{4} - \frac{x^2}{2}$
TA $= -F(x) \biggr\rvert_{-1}^{0} + F(x) \biggr\rvert_{0}^{1} - F(x) \biggr\rvert_{1}^{8}$
$=-(F(-1)-F(0))+(F(0)-F(1))-(F(8)-F(1))$
$= F(-1) - 2F(0) - 2F(1) - F(8)$
$= (\frac{1}{4}) - (2*0) + (2+\frac{1}{4}) - (-20) = \frac{83}{4}$
$\frac{83}{4}$
Area Between Curves
$\int_a^b g(x)dx + A = \int_a^b f(x)dx$
$A = \int_a^b f(x)dx - \int_a^b g(x)dx$
$= \int_a^b (f(x)-g(x)) dx$ assuming: ($f \geq g$)
$A = \int_a^b \lvert f(x) - g(x) \rvert dx$
Example 5:
Find area between $y=x^4$, $y = 8x$
$x^4 = 8x$
$x^4-8x = 0$
$x(x^3 - 8) = 0$
$x = 0, 2$
$A = \int_0^2 (8x-x^4)dx$
$= 4x^2 - \frac{1}{5}x^5 \biggr\rvert_{0}^{2}$
$= 4(2)^2-\frac{1}{5}(2)^5=16-\frac{32}{5}-0=\frac{48}{5}$
$\frac{48}{5}$
Example 6:
$y=sin(x)$, $y=cos(x)$, $[0, 2\pi]$
$sin(x) = cos(x)$
$tan(x) = 1$
$x = \frac{\pi}{4}, \frac{5\pi}{4}$
$f(x)=cos(x)-sin(x)$
$F(x)=sin(x)+cos(x)$
$A = \int_0^{2\pi} \lvert cos(x) - sin(x) \rvert dx$
$+ F(x)\biggr\rvert_{0}^{\frac{\pi}{4}} - F(x)\biggr\rvert_{\frac{\pi}{4}}^{\frac{5\pi}{4}} + F(x)\biggr\rvert_{\frac{5\pi}{4}}^{2}$
$F(\frac{\pi}{4})-F(0)-(F(\frac{\pi}{4})-F(\frac{5\pi}{4}))+(F(\frac{5\pi}{4})-F(2))$
$-F(0) + 2F(\frac{\pi}{4})-2F(\frac{5\pi}{4})+F(2\pi)$
$-1 + (2*\sqrt{2})-2(-\sqrt{2})+1$
$= 4\sqrt{2}$
$4\sqrt{2}$
Example 7:
$y=\sqrt{x}$, $y = x-2$, x-axis
$x = y^2$ and $x = y+2$
$y^2=y+2$
$y^2-y-2=0$
$(y-2)(y+1)=0$
$y = 2, -1$
$A = \int_0^2(y+2-y^2) dy$
$\frac{y^2}{2} + 2y - \frac{y^3}{3} \biggr\rvert_{0}^{2}$
$\frac{2^2}{2}+2(2)-\frac{(2)^3}{3}=2+4-\frac{8}{3}=\frac{10}{3}$
$\frac{10}{3}$