Comprehensive assessment covering all 15 lessons: antiderivatives, power rule, exponentials and logarithms, definite integrals, the Fundamental Theorem of Calculus, areas between curves, volumes of revolution, integration by substitution, integration by parts, partial fractions, differential equations, separable equations, growth and decay models, motion applications, and module synthesis.
Assessment
Select the best answer for each question. 1 mark each.
$\int (3x^2 + \frac{2}{x}) \, dx$ equals:
$\int_0^2 x^3 \, dx$ equals:
$\frac{d}{dx}\left(\int_2^{x^3} \sqrt{t} \, dt\right)$ equals:
The area between $y = x^2$ and $y = 2x$ is:
The volume when $y = x$ from $x = 0$ to $x = 3$ is rotated around the x-axis is:
$\int 2x e^{x^2} \, dx$ equals:
$\int x \cos x \, dx$ using by parts equals:
$\frac{2x+5}{(x+1)(x+2)}$ decomposes to:
The solution to $\frac{dy}{dx} = 3y$ with $y(0) = 2$ is:
A particle has $v(t) = 6t - t^2$ m/s. It changes direction when:
Short Answer
Evaluate $\int_0^1 (2x + 1)e^{x^2+x} \, dx$ using substitution. Show all working.
Find the area between $y = x^2 - 2x$ and $y = x$. Show all working.
Find the volume when the region bounded by $y = x$ and $y = x^2$ is rotated around the x-axis. Show all working.
Solve $\frac{dy}{dx} = \frac{x^2}{y^2}$ with $y(0) = 1$. Show all working.
A particle has acceleration $a(t) = 4 - 2t$ m/sยฒ. At $t = 0$, $v = 1$ m/s and $x = 0$. Find: (a) when the particle is at rest, (b) the total distance travelled from $t = 0$ to $t = 4$. Show all working.
Q1: B โ $x^3 + 2\ln|x| + C$.
Q2: C โ $[\frac{x^4}{4}]_0^2 = 4$.
Q3: B โ FTC Part 1 with chain rule: $\sqrt{x^3} \cdot 3x^2 = 3x^2\sqrt{x^3}$.
Q4: B โ Intersections at $x = 0, 2$. $\int_0^2 (2x - x^2)\,dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$.
Q5: C โ $\pi \int_0^3 x^2\,dx = \pi [\frac{x^3}{3}]_0^3 = 9\pi$.
Q6: A โ $u = x^2$, $du = 2x\,dx$. $\int e^u\,du = e^u + C = e^{x^2} + C$.
Q7: B โ $u = x$, $dv = \cos x\,dx$. $x\sin x + \cos x + C$.
Q8: B โ $2x + 5 = A(x+2) + B(x+1)$. $x = -1$: $A = 3$; $x = -2$: $B = -1$.
Q9: B โ $y = Ae^{3x}$. $y(0) = 2$: $A = 2$.
Q10: C โ $v(t) = t(6-t) = 0$ at $t = 0, 6$.
Q11 (3 marks): Let $u = x^2 + x$, $du = (2x+1)\,dx$ [1]. When $x = 0$, $u = 0$; when $x = 1$, $u = 2$ [0.5]. $\int_0^1 (2x+1)e^{x^2+x}\,dx = \int_0^2 e^u\,du$ [0.5] $= [e^u]_0^2 = e^2 - 1$ [1].
Q12 (4 marks): $x^2 - 2x = x \Rightarrow x^2 - 3x = 0 \Rightarrow x(x-3) = 0$, so $x = 0, 3$ [1]. On $[0, 3]$, $x \geq x^2 - 2x$ (test $x = 1$: $1 > -1$) [1]. $A = \int_0^3 (x - x^2 + 2x)\,dx = \int_0^3 (3x - x^2)\,dx$ [1] $= [\frac{3x^2}{2} - \frac{x^3}{3}]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}$ [1].
Q13 (4 marks): Intersections at $x = 0, 1$ [0.5]. $R(x) = x$, $r(x) = x^2$ [1]. $V = \pi \int_0^1 (x^2 - x^4)\,dx$ [1] $= \pi [\frac{x^3}{3} - \frac{x^5}{5}]_0^1 = \pi (\frac{1}{3} - \frac{1}{5}) = \frac{2\pi}{15}$ [1.5].
Q14 (4 marks): $y^2\,dy = x^2\,dx$ [1]. $\frac{y^3}{3} = \frac{x^3}{3} + C$ [1]. $y(0) = 1$: $\frac{1}{3} = 0 + C$, so $C = \frac{1}{3}$ [1]. $y^3 = x^3 + 1$, so $y = \sqrt[3]{x^3 + 1}$ [1].
Q15 (4 marks): $v(t) = 4t - t^2 + C_1$ [0.5]. $v(0) = 1$: $C_1 = 1$, so $v(t) = -t^2 + 4t + 1$ [0.5]. (a) $v(t) = 0$: $t^2 - 4t - 1 = 0$, $t = 2 + \sqrt{5} \approx 4.24$ s (only positive root in range) [1]. (b) On $[0, 4]$: $v(t) > 0$ (since $v(4) = 1 > 0$). Distance = $\int_0^4 (-t^2 + 4t + 1)\,dt = [-\frac{t^3}{3} + 2t^2 + t]_0^4$ [1] $= -\frac{64}{3} + 32 + 4 = \frac{44}{3}$ m [1].