Solutions — Area Under a Curve (Ex 11E/F/G)

Year 12 Mathematical Methods Unit 3 · Cambridge §11E/F/G · Mr Wong

ANSWER KEY

Warm-up — straight evaluations

Q1. Evaluate each definite integral:

a $\dfrac{19}{3}$ $\bigl[\tfrac{x^3}{3}\bigr]_2^3 = 9 - \tfrac{8}{3}$
b $\dfrac{1}{2}e^{8} + \dfrac{7}{2}$ $\bigl[\tfrac{1}{2}e^{2x}+x\bigr]_0^4$
c $\log_e 3$ $\bigl[\log_e(x-5)\bigr]_6^8$
d $\dfrac{1}{2}$ $\bigl[-\tfrac{1}{2}\cos(2x)\bigr]_0^{\pi/4}$

Part A — Properties of the definite integral

2 $2\!\int_1^4 f + \int_1^4 4\,dx = 2(3) + 4(3) =$ $18$
3 $\int_{-1}^2 x\,dx + 2(5) = \tfrac{3}{2} + 10 =$ $\dfrac{23}{2}$
4 $\int_{-2}^{3} f = 2+(-6)=-4$, then swap: $\displaystyle\int_3^{-2} f = -(-4) = 4$
5 $\int_0^k 8x\,dx = [4x^2]_0^k = 4k^2 = 36 \Rightarrow$ $k = 3$ (since $k>0$)

Part B — Signed vs total area

6 $\bigl[\tfrac{2}{3}x^{3/2}+x\bigr]_4^9 = (18+9)-(\tfrac{16}{3}+4) = 27 - \tfrac{28}{3} =$ $\dfrac{53}{3}$ (curve is above the axis, signed area = total area).
7 On $[1,3]$ the parabola $y=x(x-4)$ is below the axis. $\int_1^3 (x^2-4x)\,dx = -\tfrac{22}{3}$, so area $= \dfrac{22}{3}$.
8a $x$-intercept at $x=1$. $\bigl|\int_0^1 (x^2-1)\,dx\bigr| = \tfrac{2}{3}$  and  $\int_1^2 (x^2-1)\,dx = \tfrac{4}{3}$. Total area $= 2$ sq units.
8b CAS evaluated the signed integral, which is $\tfrac{4}{3} - \tfrac{2}{3} = \tfrac{2}{3}$. The negative bit on $[0,1]$ cancelled most of the positive bit on $[1,2]$. For an area you must split at $x=1$ and add absolute values.

Part C — VCAA-style

9 $\int_0^3 (-x^2+ax+12)\,dx = \bigl[-\tfrac{x^3}{3}+\tfrac{ax^2}{2}+12x\bigr]_0^3 = -9 + \tfrac{9a}{2} + 36 = \tfrac{9a}{2}+27$. Set $=45 \Rightarrow$ $a=4$. Then $-x^2+4x+12 = -(x^2-4x-12) = -(x-6)(x+2)$, giving $m=-2,\;n=6$.
10 $x$-intercept of $y=x^2-9$ at $x=3$ (inside $[0,5]$). $\int_0^3 (x^2-9)\,dx = -18$ → take abs: $18$. $\int_3^5 (x^2-9)\,dx = \tfrac{125}{3}-45+18 = \tfrac{44}{3}$. Total area $= 18 + \tfrac{44}{3} = \dfrac{98}{3}$ sq units.
Y12 Methods · §11E/F/G · Solutions Page 1 of 1 · Hand back to Mr Wong