QCE Vault / Specialist Maths Integration and applications of integration — Question 13 QCAA 2020, Paper 1 · 4 marks
Browse all questions Original practice exam Report an issue Q13 · 2020 · Technology-free Simple familiar 4 marks
QUESTION 13 (4 marks) For an exponentially distributed random variable X X X with parameter λ > 0 \lambda>0 λ > 0 ,
E ( X ) = ∫ 0 ∞ x λ e − λ x d x . E(X)=\int_0^\infty x\lambda e^{-\lambda x}\,dx. E ( X ) = ∫ 0 ∞ x λ e − λ x d x .
Use integration by parts to determine E ( X ) E(X) E ( X ) . Express your answer in simplest form. WORKED SOLUTION
QCAA guide · typeset solution 4 marks ANSWER E ( X ) = 1 λ E(X)=\frac1\lambda E ( X ) = λ 1 . Worked solution
E ( X ) = ∫ 0 ∞ x λ e − λ x d x E(X)=\int_0^\infty x\lambda e^{-\lambda x}\,dx E ( X ) = ∫ 0 ∞ x λ e − λ x d x . Take u = x u=x u = x and d v = λ e − λ x d x dv=\lambda e^{-\lambda x}dx d v = λ e − λ x d x , so d u = d x du=dx d u = d x and v = − e − λ x v=-e^{-\lambda x} v = − e − λ x . Then
E ( X ) = [ − x e − λ x ] 0 ∞ + ∫ 0 ∞ e − λ x d x = 0 + [ − e − λ x λ ] 0 ∞ = 1 λ . E(X)=\left[-xe^{-\lambda x}\right]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx=0+\left[-\frac{e^{-\lambda x}}{\lambda}\right]_0^\infty=\frac1\lambda. E ( X ) = [ − x e − λ x ] 0 ∞ + ∫ 0 ∞ e − λ x d x = 0 + [ − λ e − λ x ] 0 ∞ = λ 1 . correctly determines d u / d x du/dx d u / d x and v v v [1 mark] substitutes into the integration by parts rule
[1 mark] calculates [ − x e − λ x ] 0 ∞ [-xe^{-\lambda x}]_0^\infty [ − x e − λ x ] 0 ∞ to equal 0 [1 mark] shows that E ( X ) = 1 / λ E(X)=1/\lambda E ( X ) = 1/ λ [1 mark] One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.
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