3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Let f be continuous on [1,3] with ∫13f(x)dx=4 and ∫23f(x)dx=7. The value of ∫11/2(4f(2x)+3)dx is ______. Exactly one option is correct.
A−29
B221
C29
D215
Show answer and solution
Answer: C
Additivity: ∫12f=∫13f−∫23f=4−7=−3, so ∫21f=3. With u=2x (du=2dx; x=1→u=2, x=21→u=1): ∫11/24f(2x)dx=2∫21f(u)du=2⋅3=6. Also ∫11/23dx=3(21−1)=−23 (the constant also picks up the reversed orientation). Total 6−23=29. Forgetting the orientation gives −29; forgetting the factor 21 from du=2dx gives 12−23=221; treating ∫11/23dx as +23 gives 215.
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
The value of ∫0π/2sinx+cosxsin3xdx is ______. Exactly one option is correct.
A4π−1
B4π−2
C4π+1
D2π−1
Show answer and solution
Answer: A
The denominator is positive on [0,π/2], so the integrand is continuous. With x↦2π−x (which maps [0,2π] onto itself), I=∫0π/2cosx+sinxcos3xdx. Adding, 2I=∫0π/2sinx+cosxsin3x+cos3xdx=∫0π/2(1−sinxcosx)dx=2π−21. Hence I=4π−1.
Difficulty: Hard
Question 3Level 3 · AdvancedSingle correct
Let g:R→R be defined by g(x)=dxd(tan−1x1) for x=0 and g(0)=−1 (here tan−1 denotes the principal value). The value of ∫−31g(x)dx is ______. Exactly one option is correct.
A125π
B−125π
C127π
D−127π
Show answer and solution
Answer: D
For x=0, g(x)=1+1/x21⋅(−x21)=−1+x21, and g(0)=−1 agrees with this formula, so g(x)=−1+x21 on all of R; g is continuous. An antiderivative on the whole interval [−3,1] is −tan−1x, so the integral is −(tan−11−tan−1(−3))=−(4π+3π)=−127π. The function tan−1(1/x) is not an antiderivative on [−3,1]: it is undefined at 0 and jumps from −2π to 2π there. Using it anyway gives tan−11−tan−1(−31)=125π (option A), which even has the wrong sign although g<0. Splitting at 0 correctly: [tan−1x1]−30−+[tan−1x1]0+1=(−2π+6π)+(4π−2π)=−127π.
Difficulty: Very hard
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