Level 1 — RecognitionCalculus II — Integration

Calculus II — Integration

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1 Examination — Recognition

Time limit: 20 minutes Total marks: 30

Instructions: Answer all questions. For True/False items you must give a one-line justification to earn full marks. For matching items write the letter pairs.

Section A — Multiple Choice (1 mark each)

Q1. The most general antiderivative of $f(x)=3x^2$ is

(A) $6x + C$
(B) $x^3 + C$
(C) $x^3$
(D) $\tfrac{3}{2}x^2 + C$

Q2. $\displaystyle\int \frac{1}{x}\,dx$ equals

(A) $\ln x + C$
(B) $-x^{-2} + C$
(C) $\ln|x| + C$
(D) $\frac{x^0}{0}+C$

Q3. $\displaystyle\int \sec^2 x\,dx$ equals

(A) $\tan x + C$
(B) $\sec x\tan x + C$
(C) $-\cot x + C$
(D) $\ln|\sec x|+C$

Q4. The Fundamental Theorem of Calculus, Part 2 states that if $F'=f$ then $\int_a^b f(x)\,dx$ equals

(A) $F(b)+F(a)$
(B) $F(b)-F(a)$
(C) $f(b)-f(a)$
(D) $F'(b)-F'(a)$

Q5. For $\displaystyle\int x\,e^{x}\,dx$ the LIATE mnemonic suggests choosing $u=$

(A) $e^x$
(B) $x$
(C) $dx$
(D) $x e^x$

Q6. For $\displaystyle\int \frac{dx}{\sqrt{9-x^2}}$ the appropriate trig substitution is

(A) $x=3\sec\theta$
(B) $x=3\tan\theta$
(C) $x=3\sin\theta$
(D) $x=9\sin\theta$

Q7. The improper integral $\displaystyle\int_1^{\infty}\frac{1}{x^p}\,dx$ converges precisely when

(A) $p<1$
(B) $p\le 1$
(C) $p>1$
(D) all $p$

Q8. The volume by the disk method of the solid formed by revolving $y=f(x)$ about the $x$ -axis on $[a,b]$ is

(A) $\displaystyle\int_a^b 2\pi x f(x)\,dx$
(B) $\displaystyle\int_a^b \pi [f(x)]^2\,dx$
(C) $\displaystyle\int_a^b \pi f(x)\,dx$
(D) $\displaystyle\int_a^b [f(x)]^2\,dx$

Q9. The average value of $f$ on $[a,b]$ is

(A) $\displaystyle\int_a^b f(x)\,dx$
(B) $\dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$
(C) $\dfrac{f(a)+f(b)}{2}$
(D) $(b-a)\displaystyle\int_a^b f(x)\,dx$

Q10. The arc length of $y=f(x)$ on $[a,b]$ is

(A) $\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}\,dx$
(B) $\displaystyle\int_a^b \sqrt{1+f(x)}\,dx$
(C) $\displaystyle\int_a^b [f'(x)]^2\,dx$
(D) $\displaystyle\int_a^b 2\pi f(x)\,dx$

Section B — Matching (1 mark each pair)

Q11. Match each integral technique/form (left) with the correct partial-fraction or method template (right). (5 marks)

Left	Right
(i) $\dfrac{1}{(x-1)(x+2)}$	(A) needs $\dfrac{Ax+B}{x^2+1}$ term
(ii) $\dfrac{1}{(x-1)^2}$	(B) $\dfrac{A}{x-1}+\dfrac{B}{x+2}$
(iii) $\dfrac{1}{(x-1)(x^2+1)}$	(C) trig substitution $x=a\sec\theta$
(iv) $\sqrt{x^2-a^2}$	(D) $\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}$
(v) $\int x\ln x\,dx$	(E) integration by parts

Q12. Match the Riemann sum estimate with its sample-point rule on a partition of $[a,b]$ . (3 marks)

Left	Right
(i) Left sum	(A) uses $x_i^*=\tfrac{x_{i-1}+x_i}{2}$
(ii) Right sum	(B) uses $x_i^*=x_{i-1}$
(iii) Midpoint sum	(C) uses $x_i^*=x_i$

Section C — True / False with justification (2 marks each)

Q13. $\displaystyle\int_{-1}^{1}\frac{1}{x^2}\,dx = -2$ . (True/False + justification)

Q14. The Net Change Theorem says $\displaystyle\int_a^b F'(x)\,dx = F(b)-F(a)$ . (True/False + justification)

Q15. When using $u$ -substitution on a definite integral, if you change the limits to $u$ -values you do not need to back-substitute to $x$ . (True/False + justification)

Q16. For the area between $y=x$ and $y=x^2$ on $[0,1]$ , the integrand of the vertical-slice integral is $(x-x^2)$ . (True/False + justification)

Q17. By the comparison test, since $\dfrac{1}{x^2+1}\le \dfrac{1}{x^2}$ for $x\ge 1$ and $\int_1^\infty \frac{1}{x^2}dx$ converges, $\int_1^\infty \frac{1}{x^2+1}dx$ converges. (True/False + justification)

Q18. FTC Part 1 states $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)$ for continuous $f$ . (True/False + justification)

Answer keyMark scheme & solutions

Section A

Q1 — (B) $x^3+C$ . Since $\frac{d}{dx}(x^3+C)=3x^2$ . (1)

Q2 — (C) $\ln|x|+C$ ; absolute value needed because domain excludes $0$ and includes negatives. (1)

Q3 — (A) $\tan x + C$ , since $\frac{d}{dx}\tan x=\sec^2x$ . (1)

Q4 — (B) $F(b)-F(a)$ . (1)

Q5 — (B) $u=x$ : Logarithmic/Inverse/Algebraic/Trig/Exponential — algebraic $x$ ranks before exponential $e^x$ , so $u=x$ , $dv=e^x dx$ . (1)

Q6 — (C) $x=3\sin\theta$ ; form $\sqrt{a^2-x^2}$ with $a=3$ . (1)

Q7 — (C) $p>1$ (the $p$ -integral). (1)

Q8 — (B) $\int_a^b \pi[f(x)]^2 dx$ . (1)

Q9 — (B) $\frac{1}{b-a}\int_a^b f\,dx$ . (1)

Q10 — (A) $\int_a^b\sqrt{1+[f'(x)]^2}\,dx$ . (1)

Section B

Q11 (1 each)

(i)→(B); (ii)→(D); (iii)→(A) — wait, check: $\frac{1}{(x-1)(x^2+1)}$ decomposes as $\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$ , so (iii)→(A). (iv)→(C); (v)→(E).

Final: (i)-B, (ii)-D, (iii)-A, (iv)-C, (v)-E. (5)

Q12 (1 each)

(i)-B, (ii)-C, (iii)-A. (3)

Section C

Q13 — FALSE. The integrand $1/x^2$ has an infinite discontinuity at $x=0\in[-1,1]$ ; the integral is improper (Type II) and actually diverges. Blindly applying $[-1/x]_{-1}^{1}=-2$ is invalid. (1 for False, 1 for reason)

Q14 — TRUE. Direct statement of Net Change Theorem: integrating a rate of change $F'$ gives the total change $F(b)-F(a)$ . (2)

Q15 — TRUE. Once limits are converted to $u$ -values, the integral is fully in terms of $u$ ; evaluate directly, no back-substitution required. (2)

Q16 — TRUE. On $[0,1]$ , $x\ge x^2$ , so top minus bottom $=x-x^2\ge0$ ; vertical slices give $\int_0^1(x-x^2)\,dx$ . (2)

Q17 — TRUE. $0\le\frac{1}{x^2+1}\le\frac{1}{x^2}$ for $x\ge1$ and the larger integral converges, so by direct comparison the smaller converges. (2)

Q18 — TRUE. FTC Part 1: for continuous $f$ , $g(x)=\int_a^x f(t)\,dt$ is differentiable with $g'(x)=f(x)$ . (2)

[
  {"claim":"Antiderivative of 3x^2 is x^3","code":"x=symbols('x'); result = simplify(integrate(3*x**2,x)-x**3)==0"},
  {"claim":"Average value of f on [a,b] formula: avg of x on [0,2] equals 1","code":"x=symbols('x'); result = Rational(1,2)*integrate(x,(x,0,2))==1"},
  {"claim":"Area between y=x and y=x^2 on [0,1] equals 1/6","code":"x=symbols('x'); result = integrate(x-x**2,(x,0,1))==Rational(1,6)"},
  {"claim":"int_1^inf 1/(x^2+1) dx converges to pi/4","code":"x=symbols('x'); result = integrate(1/(x**2+1),(x,1,oo))==pi/4"},
  {"claim":"int_1^inf 1/x^2 dx converges to 1","code":"x=symbols('x'); result = integrate(1/x**2,(x,1,oo))==1"}
]