MA2100 Practice Final Exam

Winter 2026 — 60 Marks Total
Exam Structure:
Section 1: Multiple Choice — Differentiation & Integration (20 questions) — 30 marks
Section 2: Differentiation Applications (choose 2 of 5) — 10 marks
Section 3: Integration Applications (choose 4 of 5) — 20 marks

Topics covered: Transcendental functions, trig functions, laws of logarithms, exponential functions, implicit differentiation, integration formulas, u-substitution, integration by parts, integration by partial fractions, optimization, curvilinear motion, related rates, tangent/normal lines, Newton's method, area between curves, volume of solids, trapezoidal and Simpson's rules.
SECTION 1 — Differentiation & Integration (30 marks)

Find the derivative or integral as indicated. Show all work.

Derivatives (Q1–10)

1. [1.5 marks] Find the derivative of: $$y = \cos(3x^2 + 4x)$$
2. [1.5 marks] Find the derivative of: $$y = e^{2x}(e^x + e^{-x})$$
3. [1.5 marks] Find the derivative of: $$y = \log_3(\ln 2x)$$
4. [1.5 marks] Find the derivative of: $$y = \csc(\sin 3x)$$
5. [1.5 marks] Find the derivative of: $$y = \sin^{-1}(4x)$$
6. [1.5 marks] Find the derivative of: $$y = \ln(\cot 2x^3)$$
7. [1.5 marks] Find the derivative of: $$y = (\ln 5x^2)(2\sin^{-1}4x)$$
8. [1.5 marks] Find \(\dfrac{dy}{dx}\) using implicit differentiation: $$\cos(xy) + \sin 4y + e^{3x} = 4y^4$$
9. [1.5 marks] Find the derivative of: $$y = \frac{(x+1)^2}{\sec 3x}$$
10. [1.5 marks] Find the derivative of: $$y = \ln\!\big[(e^{4x})(\csc 3x^2)\big]$$

Integration (Q11–20)

11. [1.5 marks] Evaluate: $$\int x^5(x^2 - x)\,dx$$
12. [1.5 marks] Evaluate using u-substitution: $$\int \frac{3x^2}{2x^3 + 4}\,dx$$
13. [1.5 marks] Evaluate: $$\int \cos^4 4x \cdot \sin 4x\,dx$$
14. [1.5 marks] Evaluate using u-substitution: $$\int \frac{3x^2}{4x^3 + 1}\,dx$$
15. [1.5 marks] Evaluate the definite integral: $$\int_0^4 (1 - 3x^2)^2\,dx$$
16. [1.5 marks] Evaluate: $$\int \frac{3e^{5x} - 3e^x + 2}{e^{x+3}}\,dx$$
17. [1.5 marks] Evaluate using u-substitution: $$\int \frac{20\sin^{-1}2t}{\sqrt{1 - 4t^2}}\,dt$$
18. [1.5 marks] Evaluate using partial fractions: $$\int \frac{x - 19}{x^2 - 3x - 10}\,dx$$
19. [1.5 marks] Evaluate using integration by parts: $$\int x^2 \cos 2x\,dx$$
20. [1.5 marks] Evaluate the definite integral: $$\int_0^{\pi/2} (\sin 2\theta)(e^{\cos 2\theta})\,d\theta$$
SECTION 2 — Differentiation Applications (10 marks)

Choose 2 of 5 questions. Each question is worth 5 marks.

21. [5 marks] Related Rates: A ladder is slipping down along a vertical wall. If the ladder is 5.00 m long and the top is slipping at a constant rate of 4 m/s, how fast is the bottom of the ladder moving along the ground when the bottom is 2.5 m from the wall?
22. [5 marks] Related Rates: The electric resistance of a certain resistor as a function of temperature is given by \(R = 13.7 + 0.02T - 0.006T^2\) where R is in ohms and T in °C. If the temperature is increasing at 0.012 °C/s, find the rate at which the resistance is changing when \(T = 125\)°C.
23. [5 marks] Curvilinear Motion: The x- and y-coordinates of a moving particle are given by the parametric equations \(x = \sin 4t^2\) and \(y = \ln(\cos t)\), where t is in seconds and x, y in metres. Find the magnitude and direction of the resultant velocity at \(t = 3\) s.
24. [5 marks] Newton's Method: Find the positive root of \(y = 3\sin x - 1 - x^2\), when the estimate \(x_1 = 0.39\). Calculate to \(x_4\).
25. [5 marks] Tangent/Normal Lines: Find the equation of the NORMAL line to the curve \(y = \ln(\sin x)\) when \(x = \dfrac{\pi}{3}\).
Note from instructor: Problems like Q7 from old tests (trig optimization of beam dimensions) and Q8 (solving trig equations with critical values) will not be on the exam.
SECTION 3 — Integration Applications (20 marks)

Choose 4 of 5 questions. Each question is worth 5 marks.

26. [5 marks] Area Between Curves: Find the area bounded by \(y = x^2 + 2x - 8\) and \(y = x + 4\).
27. [5 marks] Area Between Curves: Find the area bounded by \(y = x^3 - x\) and the x-axis.
28. [5 marks] Volume by Disk Method: Find the volume of the solid formed by rotating the first quadrant portion of \(y = 6 - 2x\) about the x-axis.
29. [5 marks] Simpson's Rule: From an aerial photograph, a cartographer determines the widths of an island at 2.00 km intervals. The widths found are:
Distance (km)0246810121416
Width (km)03.86.29.412.110.77.53.90
Find the area of the island using Simpson's Rule.
30. [5 marks] Definite Integral Application: The rate of flow Q (in m³/s) of water over a certain dam is found by evaluating: $$Q = \int_0^{1.25} 240\sqrt{1.50 - y}\,dy$$ Find Q.
ANSWER KEY

Section 1 — Derivatives (Q1–10)

1. $$y' = -\sin(3x^2+4x)\cdot(6x+4) = -(6x+4)\sin(3x^2+4x)$$ Chain rule: derivative of cos is −sin, times derivative of inside.
2. Expand first: \(y = e^{3x} + e^{x}\). Then: $$y' = 3e^{3x} + e^{x}$$ Or use product rule on original form.
3. $$y' = \frac{1}{\ln(2x)\cdot\ln 3}\cdot\frac{1}{2x}\cdot 2 = \frac{1}{x\cdot\ln(2x)\cdot\ln 3}$$ Chain rule through log₃, then ln, then 2x.
4. $$y' = -\csc(\sin 3x)\cot(\sin 3x)\cdot 3\cos 3x$$ Chain rule: derivative of csc is −csc·cot, times derivative of sin 3x.
5. $$y' = \frac{1}{\sqrt{1-(4x)^2}}\cdot 4 = \frac{4}{\sqrt{1-16x^2}}$$
6. $$y' = \frac{1}{\cot 2x^3}\cdot(-\csc^2 2x^3)\cdot 6x^2 = \frac{-6x^2\csc^2 2x^3}{\cot 2x^3}$$
7. Product rule: \(y' = (\ln 5x^2)'\cdot(2\sin^{-1}4x) + (\ln 5x^2)\cdot(2\sin^{-1}4x)'\) $$y' = \frac{2}{x}\cdot 2\sin^{-1}4x + \ln 5x^2\cdot\frac{8}{\sqrt{1-16x^2}}$$
8. Differentiate implicitly: $$-\sin(xy)\cdot\left(y + x\frac{dy}{dx}\right) + 4\cos 4y\cdot\frac{dy}{dx} + 3e^{3x} = 16y^3\frac{dy}{dx}$$ Solve for \(\dfrac{dy}{dx}\): $$\frac{dy}{dx} = \frac{y\sin(xy) - 3e^{3x}}{16y^3 - 4\cos 4y + x\sin(xy)}$$
9. Rewrite as \(y = (x+1)^2\cos 3x\). Product rule: $$y' = 2(x+1)\cos 3x - 3(x+1)^2\sin 3x$$
10. Use log rules: \(y = 4x + \ln(\csc 3x^2)\). Then: $$y' = 4 + \frac{-\csc 3x^2\cot 3x^2\cdot 6x}{\csc 3x^2} = 4 - 6x\cot 3x^2$$

Section 1 — Integration (Q11–20)

11. Expand: \(\int (x^7 - x^6)\,dx\) $$= \frac{x^8}{8} - \frac{x^7}{7} + C$$
12. Let \(u = 2x^3+4\), \(du = 6x^2\,dx\), so \(3x^2\,dx = du/2\). $$\frac{1}{2}\int\frac{du}{u} = \frac{1}{2}\ln|2x^3+4| + C$$
13. Let \(u = \cos 4x\), \(du = -4\sin 4x\,dx\), so \(\sin 4x\,dx = -du/4\). $$-\frac{1}{4}\int u^4\,du = -\frac{1}{4}\cdot\frac{u^5}{5} + C = -\frac{\cos^5 4x}{20} + C$$
14. Let \(u = 4x^3+1\), \(du = 12x^2\,dx\), so \(3x^2\,dx = du/4\). $$\frac{1}{4}\int\frac{du}{u} = \frac{1}{4}\ln|4x^3+1| + C$$
15. Expand: \((1-3x^2)^2 = 1 - 6x^2 + 9x^4\). $$\int_0^4 (1 - 6x^2 + 9x^4)\,dx = \left[x - 2x^3 + \frac{9x^5}{5}\right]_0^4 = 4 - 128 + \frac{9216}{5} = 4 - 128 + 1843.2 = 1719.2$$
16. Rewrite by dividing each term by \(e^{x+3} = e^3\cdot e^x\): $$\int \frac{3e^{5x}}{e^{x+3}}\,dx - \int\frac{3e^x}{e^{x+3}}\,dx + \int\frac{2}{e^{x+3}}\,dx$$ $$= \int 3e^{4x-3}\,dx - \int 3e^{-3}\,dx + \int 2e^{-x-3}\,dx$$ $$= \frac{3}{4}e^{4x-3} - 3e^{-3}x - 2e^{-x-3} + C$$
17. Let \(u = \sin^{-1}2t\), \(du = \dfrac{2}{\sqrt{1-4t^2}}\,dt\), so \(\dfrac{dt}{\sqrt{1-4t^2}} = du/2\). $$20\cdot\frac{1}{2}\int u\,du = 10\cdot\frac{u^2}{2} + C = 5(\sin^{-1}2t)^2 + C$$
18. Factor: \(x^2-3x-10 = (x-5)(x+2)\). Split: \(\frac{A}{x-5}+\frac{B}{x+2}\).
Multiply: \(x-19 = A(x+2)+B(x-5)\).
Plug \(x=5\): \(-14 = 7A\) → \(A = -2\).
Plug \(x=-2\): \(-21 = -7B\) → \(B = 3\). $$-2\ln|x-5| + 3\ln|x+2| + C$$
19. Tabular method — D: \(x^2\), I: \(\cos 2x\), Signs: +, −, +
DIS
\(x^2\)\(\cos 2x\)+
\(2x\)\(\frac{1}{2}\sin 2x\)−
\(2\)\(-\frac{1}{4}\cos 2x\)+
\(0\)\(-\frac{1}{8}\sin 2x\)
$$= \frac{x^2}{2}\sin 2x + \frac{x}{2}\cos 2x - \frac{1}{4}\sin 2x + C$$
20. Let \(u = \cos 2\theta\), \(du = -2\sin 2\theta\,d\theta\).
Bounds: \(\theta=0 \to u=1\), \(\theta=\pi/2 \to u=-1\). $$-\frac{1}{2}\int_1^{-1} e^u\,du = -\frac{1}{2}[e^u]_1^{-1} = -\frac{1}{2}(e^{-1}-e^1) = \frac{e - e^{-1}}{2} \approx 1.175$$

Section 2 — Differentiation Applications (Q21–25)

21. Ladder problem. \(x^2+y^2=25\). Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\).
When \(x=2.5\): \(y=\sqrt{25-6.25}=\sqrt{18.75}\approx 4.33\).
Given \(\frac{dy}{dt}=-4\) (sliding down): $$\frac{dx}{dt} = \frac{-y\cdot dy/dt}{x} = \frac{-4.33\cdot(-4)}{2.5} = \frac{17.32}{2.5} \approx 6.93 \text{ m/s}$$
22. Resistance problem. \(\frac{dR}{dt} = \frac{dR}{dT}\cdot\frac{dT}{dt}\).
\(\frac{dR}{dT} = 0.02 - 0.012T\). At \(T=125\): \(\frac{dR}{dT} = 0.02-1.5=-1.48\). $$\frac{dR}{dt} = -1.48\times 0.012 = -0.01776 \text{ ohms/s}$$
23. Curvilinear motion.
\(v_x = \frac{dx}{dt} = \cos(4t^2)\cdot 8t\). At \(t=3\): \(v_x = 8(3)\cos(36) = 24\cos 36\).
\(v_y = \frac{dy}{dt} = \frac{-\sin t}{\cos t} = -\tan t\). At \(t=3\): \(v_y = -\tan 3\).
(Use radians.) \(\cos 36 \approx -0.905\), \(\tan 3 \approx -0.1425\).
\(v_x \approx -21.72\), \(v_y \approx 0.1425\).
Magnitude: \(|v| = \sqrt{v_x^2+v_y^2} \approx 21.72\) m/s.
Direction: \(\theta = \tan^{-1}(v_y/v_x)\).
24. Newton's method. \(f(x)=3\sin x-1-x^2\), \(f'(x)=3\cos x-2x\).
Formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\).
Starting with \(x_1=0.39\), iterate three times to find \(x_4\).
\(x_2 \approx 0.7471\), \(x_3 \approx 0.7361\), \(x_4 \approx 0.7361\).
25. Normal line. \(y = \ln(\sin x)\). \(y' = \frac{\cos x}{\sin x} = \cot x\).
At \(x=\pi/3\): slope of tangent \(= \cot(\pi/3) = 1/\sqrt{3}\).
Slope of normal \(= -\sqrt{3}\) (negative reciprocal).
\(y\)-value: \(y = \ln(\sin\pi/3) = \ln(\sqrt{3}/2) \approx -0.1438\).
Normal equation: \(y - (-0.1438) = -\sqrt{3}(x - \pi/3)\).

Section 3 — Integration Applications (Q26–30)

26. Area between curves. Set equal: \(x^2+2x-8 = x+4\) → \(x^2+x-12=0\) → \((x+4)(x-3)=0\).
Limits: \(x=-4\) to \(x=3\) (note: different from the booklet version which used \(x=-3\) to \(x=4\) — recheck your factoring).
Wait — let me recheck: \(x^2+2x-8-(x+4)=x^2+x-12=(x+4)(x-3)=0\), so \(x=-4, x=3\).
Top: \(x+4\). $$A = \int_{-4}^{3}(x+4-x^2-2x+8)\,dx = \int_{-4}^{3}(-x^2-x+12)\,dx$$ $$= \left[-\frac{x^3}{3}-\frac{x^2}{2}+12x\right]_{-4}^{3} = \frac{343}{6} \approx 57.17$$
Note: If your booklet used slightly different functions, the limits may differ. Always re-derive.
27. Area: \(y = x^3-x\) and x-axis.
Roots: \(x(x^2-1) = x(x+1)(x-1)=0\) → \(x=-1,0,1\).
Curve is above axis on \([-1,0]\), below on \([0,1]\). Split: $$A = \int_{-1}^{0}(x^3-x)\,dx + \int_0^1 -(x^3-x)\,dx$$ $$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5$$
28. Volume: \(y=6-2x\) around x-axis, first quadrant.
Limits: \(x=0\) to \(x=3\) (where \(y=0\)). $$V = \pi\int_0^3(6-2x)^2\,dx = \pi\int_0^3(36-24x+4x^2)\,dx$$ $$= \pi\left[36x-12x^2+\frac{4x^3}{3}\right]_0^3 = \pi(108-108+36) = 36\pi \approx 113.10$$
29. Simpson's Rule. \(h=2\), \(h/3 = 2/3\). Pattern: 1, 4, 2, 4, 2, 4, 2, 4, 1. $$A = \frac{2}{3}\Big[0 + 4(3.8) + 2(6.2) + 4(9.4) + 2(12.1) + 4(10.7) + 2(7.5) + 4(3.9) + 0\Big]$$ $$= \frac{2}{3}\Big[0 + 15.2 + 12.4 + 37.6 + 24.2 + 42.8 + 15.0 + 15.6 + 0\Big]$$ $$= \frac{2}{3}(162.8) = 108.5 \text{ km}^2$$
30. Dam flow rate. Let \(u = 1.50-y\), \(du = -dy\).
Bounds: \(y=0 \to u=1.50\), \(y=1.25 \to u=0.25\). $$Q = -240\int_{1.50}^{0.25} u^{1/2}\,du = -240\left[\frac{2}{3}u^{3/2}\right]_{1.50}^{0.25}$$ $$= -160\big[(0.25)^{3/2} - (1.50)^{3/2}\big] = -160[0.125 - 1.837]$$ $$= -160(-1.712) \approx 273.9 \text{ m}^3\text{/s}$$

MA2100 Practice Final — Winter 2026