Grade 12 Integrals Practice — Hard

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Question 1 of 23: ∫2/√x dx = ?

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Answer key for parents & teachers (23 questions)
  1. ∫2/√x dx = ? — 4√x + C. 2x^(−1/2): ∫2x^(−1/2) dx = 2·x^(1/2)/(1/2) + C = 4√x + C.
  2. ∫(x + 1/x)² dx = ∫(x² + 2 + 1/x²) dx = ? — x³/3 + 2x − 1/x + C. Expand: ∫x² dx + ∫2 dx + ∫x^(−2) dx = x³/3 + 2x + (−1/x) + C = x³/3 + 2x − 1/x + C.
  3. d/dx [∫₀^(x²) sin t dt] = ? (by FTC Part 1 with chain rule) — sin(x²)·2x. FTC Part 1: d/dx[∫₀^u sin t dt] = sin(u). Here u = x², so multiply by du/dx = 2x. Result: 2x sin(x²).
  4. ∫₀^2 |x − 1| dx = ? — 1. ∫₀^1 (1−x) dx + ∫₁^2 (x−1) dx = [x−x²/2]₀^1 + [x²/2−x]₁^2 = 1/2 + 1/2 = 1.
  5. ∫ x/√(x²+1) dx = ? — √(x²+1) + C. u = x²+1, du = 2x dx. ∫(1/√u)(du/2) = (1/2)·2√u + C = √(x²+1) + C.
  6. ∫ ln(x)/x dx = ? — (ln x)²/2 + C. u = ln x, du = (1/x) dx. ∫u du = u²/2 + C = (ln x)²/2 + C.
  7. Area between y = x² and y = 4 (above x² and below 4): — 32/3. Intersect: x² = 4 → x = ±2. ∫₋₂^2 (4−x²) dx = [4x−x³/3]₋₂^2 = (8−8/3)−(−8+8/3) = 2(8−8/3) = 2·16/3 = 32/3.
  8. ∫₁^2 2x(x²+1)³ dx with u=x²+1: the result is: — (5⁴−2⁴)/4 = (625−16)/4. x=1→u=2, x=2→u=5. ∫₂^5 u³ du = [u⁴/4]₂^5 = 625/4 − 16/4 = 609/4.
  9. Area between y = x and y = x³ from −1 to 1 (by symmetry or direct): — 1/2. By symmetry, 2∫₀^1 (x−x³) dx = 2[x²/2 − x⁴/4]₀^1 = 2(1/2 − 1/4) = 2(1/4) = 1/2.
  10. ∫ tan x dx = ? — −ln|cos x| + C. ∫sin x/cos x dx. u = cos x, du = −sin x dx. −∫du/u = −ln|u| + C = −ln|cos x| + C.
  11. ∫ dx/(1+x²) = ? — arctan(x) + C. ∫1/(1+x²) dx = arctan(x) + C. This is a standard formula.
  12. Area between y = eˣ and y = e^(−x) from 0 to 1: — e + 1/e − 2. ∫₀^1 (eˣ − e^(−x)) dx = [eˣ + e^(−x)]₀^1 = (e + 1/e) − (1 + 1) = e + 1/e − 2.
  13. ∫ x·√(x²+1) dx = ? — (x²+1)^(3/2)/3 + C. u = x²+1, du = 2x dx, x dx = du/2. ∫√u · du/2 = (1/2)·(2/3)u^(3/2) + C = u^(3/2)/3 + C = (x²+1)^(3/2)/3 + C.
  14. ∫ arctan x dx: u = arctan x, dv = dx: — x arctan x − (1/2)ln(1+x²) + C. u=arctan x, v=x. ∫arctan x dx = x·arctan x − ∫x/(1+x²) dx = x arctan x − (1/2)ln(1+x²) + C.
  15. ∫x²eˣ dx = ? — x²eˣ − 2xeˣ + 2eˣ + C. Double IBP: x²eˣ − ∫2xeˣ dx = x²eˣ − (2xeˣ − 2eˣ) + C = x²eˣ − 2xeˣ + 2eˣ + C.
  16. ∫eˣ sin x dx: which technique applies? — IBP applied twice (cyclic). IBP twice on ∫eˣ sin x dx cycles back to the original integral, allowing algebraic solution.
  17. For ∫eˣ cos x dx = I, after two IBP we get I = eˣ sin x + eˣ cos x − I. Solving: — I = eˣ(sin x + cos x)/2 + C. 2I = eˣ(sin x + cos x) → I = eˣ(sin x + cos x)/2 + C.
  18. ∫ x²·ln x dx (use LIATE: u = ln x): — x³ ln x/3 − x³/9 + C. u=ln x, dv=x² dx, v=x³/3. IBP: (x³/3)ln x − ∫(x³/3)(1/x) dx = (x³/3)ln x − x³/9 + C.
  19. ∫₀^1 x eˣ dx = ? — 1. F(x) = eˣ(x−1). F(1)−F(0) = e⁰ − (−1)e⁰ = 0 − (−1) = 1. Wait: F(1)=e¹(0)=0. F(0)=e⁰(−1)=−1. So 0−(−1)=1.
  20. Integration by parts CANNOT be used for: — ∫ sin(x²) dx (requires substitution, not IBP directly). ∫sin(x²) dx has no elementary antiderivative from IBP or substitution — it requires special functions.
  21. ∫ √x·ln x dx (u = ln x, dv = √x dx): — (2x^(3/2)/3)·ln x − (4x^(3/2)/9) + C. u=ln x, v=2x^(3/2)/3. IBP: (2x^(3/2)/3)ln x − ∫(2x^(3/2)/3)(1/x) dx = (2x^(3/2)/3)ln x − (2/3)∫x^(1/2) dx = (2x^(3/2)/3)ln x − (4x^(3/2)/9) + C.
  22. ∫ x³·eˣ dx (generalized): result involves: — 4 terms: x³, x², x, constant times eˣ. Each IBP reduces x power by 1: x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C — four eˣ terms.
  23. ∫ sec³x dx (a classic IBP problem) equals: — (1/2)(sec x tan x + ln|sec x + tan x|) + C. ∫sec³x dx = (1/2)(sec x tan x + ln|sec x + tan x|) + C. This is derived by IBP on ∫sec x·sec²x dx.