NCEA Level 3 Calculus 91579 Integration free practice questions with answers

Question 1 (12 marks) [Integrals and finding functions]

(a) A curve has gradient dy/dx = e^( equation of the curve, and find the y-value at x = 3 (3 significant figures).

(a) Answer: y = 2x + 2e^( • Evidence level: Merit (integrate the gradient and use a point on the curve to find the constant).

• y = ∫ (e^( • The curve passes through (0, 2): 2 = 2 + C, so C = 0.
• y = 2x + 2e^( • At x = 3: y = 15.0.

Finding a function: integrate the gradient, add C, then use the given point to find C; check by differentiating your answer.

(b) Use a suitable substitution (or recognise the reverse chain rule) to find the exact value of ∫ from 0 to 1 of 4e^(x + 2) dx. Then give the value to 3 significant figures.

(b) Answer: −4(−E + 1)e^(2) ≈ 50.8

• Evidence level: Merit (choose a substitution so the integrand becomes a standard integral).
• Let u = x + 2.
• Find an antiderivative: ∫ 4e^(x + 2) dx = 4e^(x + 2) + C.
• Evaluate between the limits: 4e^(3) − 4e^(2) = −4(−E + 1)e^(2).
• To 3 significant figures: 50.8.

Substitution: pick u so that du appears in the integrand (reverse chain rule); change the limits or go back to x, then evaluate F(b) − F(a) and show the antiderivative you used.

(c) At time t (t ≥ 0), a drug is released into the bloodstream at a rate of R = 2 + 3e^( (i) Find the total amount in the first 5 units of time (use integration; 3 significant figures). (ii) At what time has half of that total arrived? (3 significant figures)

(c) Answer: (i) 15.5; (ii) 1.99

• Evidence level: Merit (integrate a rate of change to find a total, then solve for a time).
• (i) Total = ∫ from 0 to 5 of R dt = [2x − 6e^( • (ii) We need ∫ from 0 to T of R dt = 7.75, that is 2T + 6 − 6e^(−T/2) = 7.75.
• Solving this equation numerically (a graphing calculator or solver) gives T = 1.99.

A rate integrates to a total: the area under the rate graph. To find when a given total is reached, set the integral from 0 to T equal to that total and solve for T. x