Level 2 — RecallCalculus II — Integration

Calculus II — Integration

30 minutes40 marksprintable — key stays hidden on paper

Level 2 Test Paper — Recall & Standard Problems

Time limit: 30 minutes Total marks: 40

Instructions: Answer all questions. Show all working. Use ...... notation for mathematics. Include the constant of integration where appropriate.


Question 1. (3 marks) State the definition of an antiderivative of a function ff, and explain why the general antiderivative includes a constant +C+C.

Question 2. (4 marks) Evaluate the following indefinite integrals: (a) (3x24x+2x)dx\displaystyle \int \left(3x^2 - 4x + \frac{2}{x}\right)\,dx (2 marks) (b) (ex+sec2x)dx\displaystyle \int \left(e^{x} + \sec^2 x\right)\,dx (2 marks)

Question 3. (5 marks) Use the substitution u=x2+1u = x^2 + 1 to evaluate the definite integral 02x(x2+1)3dx,\int_{0}^{2} x\,(x^2+1)^3\,dx, remembering to change the limits of integration.

Question 4. (5 marks) Use integration by parts to evaluate xexdx.\int x\,e^{x}\,dx. State clearly your choices of uu and dvdv (mention LIATE).

Question 5. (5 marks) State both parts of the Fundamental Theorem of Calculus. Then, using Part 1, compute ddx1x2sintdt.\frac{d}{dx}\int_{1}^{x^2} \sin t\,dt.

Question 6. (4 marks) Evaluate the trigonometric integral sin3xcos2xdx.\int \sin^3 x \cos^2 x\,dx.

Question 7. (5 marks) Decompose into partial fractions and integrate: 5x4x2x2dx.\int \frac{5x-4}{x^2-x-2}\,dx.

Question 8. (4 marks) Determine whether the improper integral converges, and if so find its value: 11x2dx.\int_{1}^{\infty} \frac{1}{x^2}\,dx.

Question 9. (5 marks) Find the area of the region enclosed between the curves y=x2y = x^2 and y=2xy = 2x.


End of paper

Answer keyMark scheme & solutions

Question 1. (3 marks)

  • An antiderivative of ff on an interval II is a function FF such that F(x)=f(x)F'(x)=f(x) for all xIx\in I. (1)
  • If FF and GG are both antiderivatives, then (FG)=ff=0(F-G)'=f-f=0, so FGF-G is constant (by the Mean Value Theorem / a function with zero derivative is constant). (1)
  • Hence every antiderivative differs from FF by a constant, so the general form is F(x)+CF(x)+C. (1)

Question 2. (4 marks) (a) (3x24x+2/x)dx=x32x2+2lnx+C\int(3x^2-4x+2/x)\,dx = x^3 - 2x^2 + 2\ln|x| + C. (2) (1 for power terms, 1 for log term) (b) (ex+sec2x)dx=ex+tanx+C\int(e^x + \sec^2 x)\,dx = e^x + \tan x + C. (2)


Question 3. (5 marks)

  • u=x2+1du=2xdxxdx=12duu=x^2+1 \Rightarrow du = 2x\,dx \Rightarrow x\,dx = \tfrac12 du. (1)
  • Limits: x=0u=1x=0\Rightarrow u=1; x=2u=5x=2\Rightarrow u=5. (1)
  • Integral =1215u3du=12[u44]15=\tfrac12\int_1^5 u^3\,du = \tfrac12\cdot\left[\tfrac{u^4}{4}\right]_1^5. (1)
  • =18(6251)=6248=78=\tfrac18(625-1)=\tfrac{624}{8}=78. (2)

Question 4. (5 marks)

  • By LIATE, let u=xu=x (algebraic), dv=exdxdv=e^x dx. Then du=dxdu=dx, v=exv=e^x. (2)
  • xexdx=xexexdx\int x e^x dx = xe^x - \int e^x dx. (2)
  • =xexex+C=ex(x1)+C= xe^x - e^x + C = e^x(x-1)+C. (1)

Question 5. (5 marks)

  • FTC Part 1: If g(x)=axf(t)dtg(x)=\int_a^x f(t)\,dt with ff continuous, then g(x)=f(x)g'(x)=f(x). (1)
  • FTC Part 2: If ff is continuous and FF is any antiderivative, then abf(x)dx=F(b)F(a)\int_a^b f(x)\,dx = F(b)-F(a). (1)
  • Let G(x)=1x2sintdtG(x)=\int_1^{x^2}\sin t\,dt. With u=x2u=x^2, ddx1usintdt=sin(u)dudx\frac{d}{dx}\int_1^{u}\sin t\,dt = \sin(u)\cdot\frac{du}{dx} (chain rule). (2)
  • =sin(x2)2x=2xsin(x2)=\sin(x^2)\cdot 2x = 2x\sin(x^2). (1)

Question 6. (4 marks)

  • Odd power of sine: write sin3x=sinx(1cos2x)\sin^3 x = \sin x(1-\cos^2 x). (1)
  • sinx(1cos2x)cos2xdx\int \sin x (1-\cos^2 x)\cos^2 x\,dx. Let u=cosxu=\cos x, du=sinxdxdu=-\sin x\,dx. (1)
  • =(1u2)u2du=(u2u4)du=u33+u55+C= -\int (1-u^2)u^2\,du = -\int(u^2-u^4)\,du = -\tfrac{u^3}{3}+\tfrac{u^5}{5}+C. (1)
  • =cos5x5cos3x3+C= \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C. (1)

Question 7. (5 marks)

  • Factor: x2x2=(x2)(x+1)x^2-x-2=(x-2)(x+1). (1)
  • 5x4(x2)(x+1)=Ax2+Bx+1\frac{5x-4}{(x-2)(x+1)} = \frac{A}{x-2}+\frac{B}{x+1}. Then 5x4=A(x+1)+B(x2)5x-4 = A(x+1)+B(x-2).
  • x=2x=2: 6=3AA=26=3A\Rightarrow A=2. x=1x=-1: 9=3BB=3-9=-3B\Rightarrow B=3. (2)
  • Integral =(2x2+3x+1)dx=2lnx2+3lnx+1+C=\int\left(\frac{2}{x-2}+\frac{3}{x+1}\right)dx = 2\ln|x-2|+3\ln|x+1|+C. (2)

Question 8. (4 marks)

  • 1x2dx=limb1bx2dx\int_1^\infty x^{-2}dx = \lim_{b\to\infty}\int_1^b x^{-2}dx. (1)
  • =limb[1x]1b=limb(1b+1)=\lim_{b\to\infty}\left[-\tfrac1x\right]_1^b = \lim_{b\to\infty}\left(-\tfrac1b + 1\right). (2)
  • =1= 1. Converges to 11. (1)

Question 9. (5 marks)

  • Intersections: x2=2xx=0,2x^2=2x \Rightarrow x=0, 2. (1)
  • On [0,2][0,2], 2xx22x \ge x^2, so area =02(2xx2)dx=\int_0^2 (2x-x^2)\,dx. (1)
  • =[x2x33]02=483=\left[x^2-\tfrac{x^3}{3}\right]_0^2 = 4-\tfrac83. (2)
  • =43=\frac{4}{3}. (1)

[
  {"claim":"Q3: definite integral equals 78","code":"x=symbols('x'); result = integrate(x*(x**2+1)**3,(x,0,2))==78"},
  {"claim":"Q7: partial fraction integral has coefficients 2 and 3","code":"x=symbols('x'); A,B=2,3; expr=simplify(A*(x+1)+B*(x-2)-(5*x-4)); result = expr==0"},
  {"claim":"Q8: improper integral converges to 1","code":"x=symbols('x'); result = integrate(1/x**2,(x,1,oo))==1"},
  {"claim":"Q9: area between curves equals 4/3","code":"x=symbols('x'); result = integrate(2*x-x**2,(x,0,2))==Rational(4,3)"}
]