QCE Vault / Mathematical Methods Integration practice QCE Mathematical Methods · Original practice questions with worked solutions
Browse all questions Differentiation practice Integration practice Original practice exam Practise antiderivatives, definite integrals and area calculations. Look for a standard form or a numerator related to the derivative of the denominator before doing unnecessary algebra.
Key ideas An indefinite integral gives a family of antiderivatives, so include the constant C C C . The fundamental theorem of calculus gives ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx=F(b)-F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) when F ′ = f F'=f F ′ = f and the hypotheses hold. A definite integral gives signed area. For total area, split at sign changes; for area between curves, integrate upper minus lower.
Worked example ∫ 0 2 ( 3 x 2 + 1 ) d x = [ x 3 + x ] 0 2 = 10 \int_0^2(3x^2+1)\,dx=[x^3+x]_0^2=10 ∫ 0 2 ( 3 x 2 + 1 ) d x = [ x 3 + x ] 0 2 = 10 . Differentiating x 3 + x x^3+x x 3 + x recovers the integrand. A common mistake Do not report signed area as total geometric area when a curve crosses the horizontal axis.
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Q4 · Practice question Technology-free Simple familiar 1 mark
QUESTION 4 An antiderivative of 6 x 2 − 4 x + 1 6x^2-4x+1 6 x 2 − 4 x + 1 is (A) 6 x 3 − 4 x 2 + x + C 6x^3-4x^2+x+C 6 x 3 − 4 x 2 + x + C (B) 3 x 2 − 2 x + C 3x^2-2x+C 3 x 2 − 2 x + C (C) 2 x 3 − 4 x 2 + x + C 2x^3-4x^2+x+C 2 x 3 − 4 x 2 + x + C (D) 2 x 3 − 2 x 2 + x + C 2x^3-2x^2+x+C 2 x 3 − 2 x 2 + x + C WORKED SOLUTION
Answer D 1 mark Worked solution
Integrating term-by-term gives 2 x 3 − 2 x 2 + x + C 2x^3-2x^2+x+C 2 x 3 − 2 x 2 + x + C . Selects the correct option.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus Q29 · Practice question Technology-free Simple familiar 2 marks
QUESTION 29 (2 marks) Evaluate ∫ 1 3 ( 2 x + 1 ) d x \displaystyle\int_1^3(2x+1)\,dx ∫ 1 3 ( 2 x + 1 ) d x . WORKED SOLUTION
Practice marking scheme 2 marks Worked solution
∫ ( 2 x + 1 ) d x = x 2 + x \int(2x+1)dx=x^2+x ∫ ( 2 x + 1 ) d x = x 2 + x . Hence [ x 2 + x ] 1 3 = 12 − 2 = 10 [x^2+x]_1^3=12-2=10 [ x 2 + x ] 1 3 = 12 − 2 = 10 . Finds an antiderivative x 2 + x x^2+x x 2 + x . [1 mark] Evaluates the definite integral as 10 10 10 . [1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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Q31 · Practice question Technology-free Simple familiar 3 marks
QUESTION 31 (3 marks) Evaluate ∫ 0 1 4 x x 2 + 1 d x \displaystyle\int_0^1\frac{4x}{x^2+1}\,dx ∫ 0 1 x 2 + 1 4 x d x . WORKED SOLUTION
Practice marking scheme 3 marks Worked solution
Let u = x 2 + 1 u=x^2+1 u = x 2 + 1 , so d u = 2 x d x du=2x\,dx d u = 2 x d x . The integral becomes 2 ∫ 1 2 d u / u = 2 ln 2 2\int_1^2du/u=2\ln2 2 ∫ 1 2 d u / u = 2 ln 2 . Recognises or applies the substitution u = x 2 + 1 u=x^2+1 u = x 2 + 1 . [1 mark] Obtains 2 ln ( x 2 + 1 ) 2\ln(x^2+1) 2 ln ( x 2 + 1 ) as an antiderivative. [1 mark] Evaluates the bounds to obtain 2 ln 2 2\ln2 2 ln 2 . [1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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All integration practice questions 73 original questions · Page 1 of 4
Q4 · Original practice · 1 mark An antiderivative of 6 x 2 − 4 x + 1 6x^2-4x+1 6 x 2 − 4 x + 1 is Introduction to integration Q6 · Original practice · 1 mark Evaluate ∫ 0 2 3 x 2 d x \displaystyle\int_0^2 3x^2\,dx ∫ 0 2 3 x 2 d x . Further integration Q14 · Original practice · 1 mark A particle has velocity v ( t ) = 3 t 2 − 4 t + 2 v(t)=3t^2-4t+2 v ( t ) = 3 t 2 − 4 t + 2 . Its displacement from t = 0 t=0 t = 0 to t = 2 t=2 t = 2 is Introduction to integration Q16 · Original practice · 1 mark An antiderivative of 2 x x 2 + 5 \dfrac{2x}{x^2+5} x 2 + 5 2 x is Further integration Q24 · Original practice · 1 mark A differentiable function satisfies f ′ ( x ) = 3 x 2 − 2 f'(x)=3x^2-2 f ′ ( x ) = 3 x 2 − 2 and f ( 1 ) = 4 f(1)=4 f ( 1 ) = 4 . Determine f ( 2 ) f(2) f ( 2 ) . Introduction to integration Q29 · Original practice · 2 marks Evaluate ∫ 1 3 ( 2 x + 1 ) d x \displaystyle\int_1^3(2x+1)\,dx ∫ 1 3 ( 2 x + 1 ) d x . Introduction to integration Q31 · Original practice · 3 marks Evaluate ∫ 0 1 4 x x 2 + 1 d x \displaystyle\int_0^1\frac{4x}{x^2+1}\,dx ∫ 0 1 x 2 + 1 4 x d x . Further integration Q39 · Original practice · 4 marks A particle moves along a straight line with velocity v ( t ) = t 2 − 4 t + 3 v(t)=t^2-4t+3 v ( t ) = t 2 − 4 t + 3 for 0 ≤ t ≤ 4 0\le t\le4 0 ≤ t ≤ 4 . Determine the total distance travelled. Introduction to integration Q41 · Original practice · 3 marks For 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 , the graph y = k x ( 1 − x ) y=kx(1-x) y = k x ( 1 − x ) lies above the x x x -axis. The area between the graph and the x x x -axis is 2 square units. Determine k k k . Further integration Q46 · Original practice · 4 marks The curves y = x 2 y=x^2 y = x 2 and y = k x y=kx y = k x , where k > 0 k>0 k > 0 , enclose a bounded region of area 9 2 \dfrac92 2 9 . Determine k k k . Further integration Q54 · Original practice · 1 mark An antiderivative of 4 x 3 + 2 4x^3+2 4 x 3 + 2 is Introduction to integration Q56 · Original practice · 1 mark Evaluate ∫ 0 1 e 2 x d x \displaystyle\int_0^1 e^{2x}\,dx ∫ 0 1 e 2 x d x . Further integration Q64 · Original practice · 1 mark A particle has velocity v ( t ) = 2 t − 1 v(t)=2t-1 v ( t ) = 2 t − 1 . Its displacement from t = 0 t=0 t = 0 to t = 3 t=3 t = 3 is Introduction to integration Q66 · Original practice · 1 mark An antiderivative of 6 x 3 x 2 + 4 \dfrac{6x}{3x^2+4} 3 x 2 + 4 6 x is Further integration Q74 · Original practice · 1 mark A function satisfies f ′ ( x ) = 2 x + 3 f'(x)=2x+3 f ′ ( x ) = 2 x + 3 and ∫ 0 2 f ( x ) d x = 12 \displaystyle\int_0^2 f(x)\,dx=12 ∫ 0 2 f ( x ) d x = 12 . Determine f ( 0 ) f(0) f ( 0 ) . Introduction to integration Q79 · Original practice · 2 marks Evaluate ∫ 1 4 ( 3 x + 2 ) d x \displaystyle\int_1^4(3x+2)\,dx ∫ 1 4 ( 3 x + 2 ) d x . Introduction to integration Q81 · Original practice · 2 marks Evaluate ∫ 1 3 2 x x 2 + 1 d x \displaystyle\int_1^3\frac{2x}{x^2+1}\,dx ∫ 1 3 x 2 + 1 2 x d x . Further integration Q89 · Original practice · 5 marks A particle has acceleration a ( t ) = 6 t − 12 a(t)=6t-12 a ( t ) = 6 t − 12 , velocity v ( 0 ) = 9 v(0)=9 v ( 0 ) = 9 and position s ( 0 ) = 4 s(0)=4 s ( 0 ) = 4 . Determine the total distance travelled for 0 ≤ t ≤ 4 0\le t\le4 0 ≤ t ≤ 4 . Introduction to integration Q91 · Original practice · 4 marks Determine the area enclosed by the curves y = 4 − x 2 y=4-x^2 y = 4 − x 2 and y = x + 2 y=x+2 y = x + 2 . Further integration Q96 · Original practice · 5 marks The curves y = k − x 2 y=k-x^2 y = k − x 2 and y = x y=x y = x , where k > 0 k>0 k > 0 , enclose a region of area 9 2 \dfrac92 2 9 square units, as shown. Determine k k k . Further integration Q116 · Original practice · 1 mark An antiderivative of 5 e 2 x − 3 x 5e^{2x}-\dfrac{3}{x} 5 e 2 x − x 3 , for x > 0 x>0 x > 0 , is Introduction to integration Q117 · Original practice · 1 mark A function satisfies f ′ ( x ) = 4 x 3 − 2 f'(x)=4x^3-2 f ′ ( x ) = 4 x 3 − 2 and f ( 1 ) = 5 f(1)=5 f ( 1 ) = 5 . Determine f ( 0 ) f(0) f ( 0 ) . Introduction to integration Q118 · Original practice · 1 mark A particle has acceleration a ( t ) = 6 t − 4 a(t)=6t-4 a ( t ) = 6 t − 4 , with v ( 0 ) = 3 v(0)=3 v ( 0 ) = 3 and s ( 0 ) = 1 s(0)=1 s ( 0 ) = 1 . Determine s ( 2 ) s(2) s ( 2 ) . Introduction to integration Q119 · Original practice · 3 marks A particle moves along a straight line with velocity v ( t ) = 4 − 2 t v(t)=4-2t v ( t ) = 4 − 2 t and initial position s ( 0 ) = 3 s(0)=3 s ( 0 ) = 3 . Determine s ( t ) s(t) s ( t ) and the first time after t = 0 t=0 t = 0 at which the particle returns to its initial position. Introduction to integration