Covers Lessons 6โ10: areas between curves, volumes of revolution, integration by substitution, integration by parts, and partial fractions.
Assessment
Select the best answer for each question.
The area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$ is:
The volume when $y = x$ from $x = 0$ to $x = 2$ is rotated around the x-axis is:
$\int 2x \cos(x^2) \, dx$ equals:
For $\int x e^x \, dx$, the LIATE rule suggests choosing:
$\frac{3}{(x-1)(x+2)}$ decomposes to:
The washer method is used when:
$\int_0^1 x e^{x^2} \, dx$ equals:
$\int x \ln x \, dx$ using by parts equals:
Short Answer
Find the area between $y = x^2$ and $y = 2x + 3$. Show all working.
Find the volume when $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated around the x-axis. Show all working.
Evaluate $\int_0^{\ln 2} e^x \sqrt{1 + e^x} \, dx$ using substitution. Show all working.
Find $\int x e^{2x} \, dx$ using integration by parts. Show all working.
Q1: B โ $\int_0^1 (x - x^2) \, dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.
Q2: B โ $V = \pi \int_0^2 x^2 \, dx = \pi [\frac{x^3}{3}]_0^2 = \frac{8\pi}{3}$ (volume of cone).
Q3: B โ $u = x^2$, $du = 2x \, dx$, so $\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C$.
Q4: B โ LIATE: Algebraic before Exponential, so $u = x$.
Q5: B โ $3 = A(x+2) + B(x-1)$. $x = 1$: $A = 1$; $x = -2$: $B = -1$.
Q6: A โ Washer method handles regions with a gap (hole) from the axis.
Q7: B โ $u = x^2$, $\frac{1}{2}\int_0^1 e^u \, du = \frac{1}{2}[e^u]_0^1 = \frac{1}{2}(e-1)$.
Q8: B โ $u = \ln x$, $dv = x \, dx$. $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$.
Q9 (4 marks): $x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0$, so $x = -1, 3$ [1]. On $[-1, 3]$, $2x + 3 \geq x^2$ [0.5]. $A = \int_{-1}^{3} (2x + 3 - x^2) \, dx = [x^2 + 3x - \frac{x^3}{3}]_{-1}^{3}$ [1] $= (9 + 9 - 9) - (1 - 3 + \frac{1}{3}) = 9 + \frac{5}{3} = \frac{32}{3}$ [1.5].
Q10 (4 marks): $V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$ [1] $= \pi [\frac{x^2}{2}]_0^4$ [1] $= \pi \cdot 8 = 8\pi$ [2].
Q11 (4 marks): Let $u = 1 + e^x$, $du = e^x \, dx$ [1]. When $x = 0$, $u = 2$; when $x = \ln 2$, $u = 3$ [1]. $\int_0^{\ln 2} e^x \sqrt{1+e^x} \, dx = \int_2^3 \sqrt{u} \, du$ [1] $= [\frac{2}{3}u^{3/2}]_2^3 = \frac{2}{3}(3\sqrt{3} - 2\sqrt{2})$ [1].
Q12 (4 marks): $u = x$, $dv = e^{2x} \, dx$, so $du = dx$, $v = \frac{1}{2}e^{2x}$ [1]. $\int x e^{2x} \, dx = \frac{x}{2}e^{2x} - \frac{1}{2}\int e^{2x} \, dx$ [1] $= \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$ [1] $= \frac{e^{2x}}{4}(2x - 1) + C$ [1].