Grade 12 Integrals Practice

Question 1 of 45Score 0/0Medium

Question 1 of 45: ∫1/x dx = ?

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Keep going: Read the lesson: Antiderivatives & Indefinite Integrals Integral Evaluator
Answer key for parents & teachers (45 questions)
  1. ∫1/x dx = ? — ln|x| + C. ∫1/x dx = ln|x| + C. The absolute value handles negative x values.
  2. ∫x^(−2) dx = ? — −x^(−1) + C. ∫x^(−2) dx = x^(−1)/(−1) + C = −x^(−1) + C = −1/x + C.
  3. ∫√x dx = ? — (2/3)x^(3/2) + C. √x = x^(1/2). ∫x^(1/2) dx = x^(3/2)/(3/2) + C = (2/3)x^(3/2) + C.
  4. ∫(eˣ + 1/x) dx = ? — eˣ + ln|x| + C. ∫eˣ dx + ∫1/x dx = eˣ + ln|x| + C.
  5. ∫(4x³ − 6x) dx = ? — x⁴ − 3x² + C. ∫4x³ dx = x⁴. ∫6x dx = 3x². Total: x⁴ − 3x² + C.
  6. ∫(sin x + cos x) dx = ? — −cos x + sin x + C. ∫sin x = −cos x; ∫cos x = sin x. Total: −cos x + sin x + C.
  7. The indefinite integral ∫f(x) dx represents: — A family of functions (all antiderivatives). ∫f(x) dx = F(x) + C — the entire family of antiderivatives, parameterized by the constant C.
  8. ∫sec²x dx = ? — tan x + C. ∫sec²x dx = tan x + C. (Verify: d/dx[tan x] = sec²x. ✓)
  9. Initial condition problem: if F'(x) = 2x and F(0) = 3, then F(x) = ? — x² + 3. ∫2x dx = x² + C. Apply F(0) = 3: 0 + C = 3 → C = 3. F(x) = x² + 3.
  10. FTC Part 1 states: d/dx ∫_a^x f(t) dt = ? — f(x). FTC Part 1: the derivative of the integral with variable upper limit equals the integrand evaluated at x: d/dx ∫_a^x f(t) dt = f(x).
  11. ∫₀^π sin x dx = ? — 2. F(x) = −cos x. F(π) − F(0) = −cos π − (−cos 0) = −(−1) − (−1) = 1 + 1 = 2.
  12. ∫₀^1 eˣ dx = ? — e − 1. F(x) = eˣ. F(1) − F(0) = e¹ − e⁰ = e − 1.
  13. ∫_a^b f(x) dx = −∫_b^a f(x) dx. This means: — Swapping limits negates the integral. Swapping the limits of integration changes the sign: ∫_a^b f dx = −∫_b^a f dx.
  14. A Riemann sum approximates ∫_a^b f(x) dx by: — Summing n rectangle areas with width Δx = (b−a)/n. Riemann sum: Σ f(xᵢ*)·Δx. As n→∞, this converges to the exact definite integral.
  15. ∫₀^2 (3x² − 2x) dx = ? — 4. F(x) = x³ − x². F(2) = 8 − 4 = 4. F(0) = 0. Integral = 4.
  16. If f(x) < 0 on [a,b], then ∫_a^b f(x) dx is: — Negative. When f is negative on the entire interval, the net signed area is negative.
  17. The average value of f on [a,b] is: — (1/(b−a))·∫_a^b f(x) dx. Average value = (1/(b−a))·∫_a^b f(x) dx. This divides total area by the length of the interval.
  18. ∫₁^4 1/x dx = ? — ln 4 − ln 1 = ln 4. F(x) = ln|x|. F(4) − F(1) = ln 4 − ln 1 = ln 4 − 0 = ln 4 ≈ 1.386.
  19. The net area ∫₋₁^1 x dx = ? — 0. F(x) = x²/2. F(1) − F(−1) = 1/2 − 1/2 = 0. The positive area [0,1] and negative area [−1,0] cancel.
  20. If ∫₀^3 f(x) dx = 7 and ∫₀^1 f(x) dx = 2, then ∫₁^3 f(x) dx = ? — 5. By additivity: ∫₀^3 = ∫₀^1 + ∫₁^3. So ∫₁^3 = 7 − 2 = 5.
  21. ∫₋π^π sin x dx = ? — 0. sin x is an odd function on a symmetric interval [−π, π]. ∫₋a^a (odd function) dx = 0.
  22. ∫₋π^π cos x dx = ? — 0. F(x) = sin x. F(π) − F(−π) = sin π − sin(−π) = 0 − 0 = 0.
  23. Evaluate: ∫₀^1 (x + 1)^2 dx — 7/3. F(x) = (x+1)³/3. F(1) = 8/3. F(0) = 1/3. Integral = 8/3 − 1/3 = 7/3.
  24. The number that represents the exact signed area between y = x² and the x-axis from 0 to 3 is: — 27/3 = 9. ∫₀^3 x² dx = [x³/3]₀^3 = 27/3 − 0 = 9. Options A and D are equivalent.
  25. ∫3x²(x³+1)⁴ dx = ? — (x³+1)⁵/5 + C. u = x³+1, du = 3x² dx. ∫u⁴ du = u⁵/5 + C = (x³+1)⁵/5 + C.
  26. ∫2xe^(x²) dx = ? — e^(x²) + C. u = x², du = 2x dx. ∫eᵘ du = eᵘ + C = e^(x²) + C.
  27. ∫cos(3x) dx = ? — sin(3x)/3 + C. u = 3x, du = 3 dx, so dx = du/3. ∫cos(u)(du/3) = (1/3)sin(u) + C = sin(3x)/3 + C.
  28. ∫sin(x²)·2x dx = ? — −cos(x²) + C. u = x², du = 2x dx. ∫sin(u) du = −cos(u) + C = −cos(x²) + C.
  29. The area between f(x) = x and g(x) = x² from 0 to 1 is: — 1/6. f ≥ g on [0,1]. ∫₀^1 (x−x²) dx = [x²/2 − x³/3]₀^1 = 1/2 − 1/3 = 1/6.
  30. To find area between two curves, we need to know where they intersect because: — The limits of integration are the intersection points. The intersection points determine the limits of integration [a,b] over which to compute ∫[f(x)−g(x)] dx.
  31. ∫e^(5x) dx = ? — e^(5x)/5 + C. u = 5x, du = 5 dx. ∫eᵘ du/5 = eᵘ/5 + C = e^(5x)/5 + C.
  32. Area between y = sin x and x-axis from 0 to π: — 2. sin x ≥ 0 on [0,π]. ∫₀^π sin x dx = [−cos x]₀^π = −cos π + cos 0 = 1 + 1 = 2.
  33. For definite integrals with u-sub, we can convert limits. For ∫₀^1 2x(x²+1)³ dx with u=x²+1: — New limits: u(0)=1 to u(1)=2. u = x² + 1: when x=0, u=1; when x=1, u=2. New integral: ∫₁^2 u³ du.
  34. Which integral requires a different method than u-substitution? — ∫x·sin(x) dx. ∫x sin(x) dx is a product of two unrelated functions — it requires integration by parts, not u-substitution.
  35. ∫ 2x/(x²+3) dx = ? — ln(x²+3) + C. u = x²+3, du = 2x dx. ∫du/u = ln|u| + C = ln(x²+3) + C.
  36. ∫x eˣ dx: using IBP with u = x, dv = eˣ dx gives: — xeˣ − eˣ + C. u=x, v=eˣ, du=dx. ∫x eˣ dx = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C.
  37. ∫x sin x dx: using u=x, dv=sin x dx: — −x cos x + sin x + C. u=x, v=−cos x. ∫x sin x dx = x·(−cos x) − ∫(−cos x) dx = −x cos x + sin x + C.
  38. ∫ln x dx: use u = ln x, dv = dx: — x ln x − x + C. u=ln x, v=x. ∫ln x dx = x·ln x − ∫x·(1/x) dx = x ln x − ∫1 dx = x ln x − x + C.
  39. For ∫x²eˣ dx, IBP must be applied: — Twice. Each IBP reduces the polynomial power by 1. x² needs to reach x⁰ (constant), requiring 2 applications.
  40. Integration by parts is derived from the: — Product rule d(uv) = u dv + v du. d(uv) = u dv + v du. Integrate both sides: uv = ∫u dv + ∫v du. Rearrange: ∫u dv = uv − ∫v du.
  41. ∫x cos x dx = ? — x sin x + cos x + C. u=x, dv=cos x dx, v=sin x. ∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x + C.
  42. When should you choose dv = eˣ dx rather than u = eˣ? — When no LIATE option is higher priority than exponential. By LIATE, eˣ is last priority for u. So eˣ becomes dv unless the other factor is also exponential (same priority).
  43. ∫ x^n·eˣ dx requires n IBP applications. For n=4: — 4 applications. Each IBP reduces the power of x by 1. For x⁴, we need 4 applications to reach x⁰.
  44. The "cyclic trick" in IBP occurs when: — The second IBP reproduces the original integral. For ∫eˣ sin x dx, two IBP steps return −I, giving I = ∫eˣ sin x dx = eˣ(sin x − cos x)/2 + C.
  45. Which integral can be solved by u-sub alone (no IBP)? — ∫x·e^(x²) dx. ∫x·e^(x²) dx: u = x², du = 2x dx. Direct u-sub works. The others need IBP.