Four linked problems drawn from this topic’s subtopics. 12 parts · untimed · answers and worked solutions at the end.
Example 1
The function f is continuous on [−4,3]. The table gives exact signed integrals; no formula for f is supplied.
Find ∫−13(f(x))dx.
Given signed integrals
Integral
Value
∫−4−1(f(x))dx
3
∫3−4(f(x))dx
−5
Show worked answer
Answer: 2
Reverse the second given integral: ∫−43(f(x))dx=5.
By additivity, the required integral is 5−(3)=2.
Example 2
Angles are in radians. F′(x)=−25cos(5x)−5Bsin(5x), where B is unknown. Also F(0)=5 and F(5π)=−5.
a Find B.
Find F(10π).
Show worked answer
Answer: −5
The initial condition gives C=5−(5)=0.
At x=10π,sin(5x)=1 and cos(5x)=0.
Therefore F(10π)=−5+0=−5.
Example 3
The graph of f consists exactly of the straight segments joining the labelled points. Use the four shown subintervals of width 3. Estimates L and U refer to the signed integral, so a rectangle below the axis has negative signed contribution.
a Calculate the lower sum L for ∫012(f(x))dx.
b Calculate the upper sum U using the same four subintervals.
Find the total geometric area between the graph and the x−axis.
f(x)
036912-5-2.502.5ABCDE
x
A: (0,2)B: (3,0)C: (6,−5)D: (9,0)E: (12,2)
fShow worked answer
Answer: 21
The two end triangles above the axis have combined area (2+2)×23=6.
The two central triangles below the axis have combined geometric area 2×3×25=15.
Add positive areas: 6+15=21 square units.
By contrast, 2L+U=−9 is the signed integral, not the total geometric area.
Integration Review Practice | Year 12 Methods | Ascent Arcade