Integral Calculus
GIF inside definite integral
MMTS_Full_Test_22
Grade 12
Question:
Let $I=\displaystyle\int_0^2\!\left[\left|x^2-5x+4\right|+\left[\sin\frac{3\pi}{2}x\right]\right]dx$ (where $[\cdot]$ is GIF). Then $I+\dfrac{2}{3}$ is
(A) $\dfrac{13-\sqrt{17}-\sqrt{21}-2\sqrt{5}-\sqrt{13}}{2}$
(B) $\dfrac{13+\sqrt{17}-\sqrt{21}-\sqrt{5}-\sqrt{13}}{2}$
(C) $\dfrac{14-\sqrt{17}-\sqrt{21}+\sqrt{5}-\sqrt{13}}{2}$
(D) $\dfrac{14-\sqrt{17}-\sqrt{21}+2\sqrt{5}-\sqrt{13}}{2}$
Step-by-Step Solution
Key Concept: Evaluate $[\sin(3\pi x/2)]$ on each subinterval of $[0,2]$ to get the piecewise-constant term. Then integrate $|x^2-5x+4|$ after splitting at $x=1$ (root in $[0,2]$).
After splitting at GIF boundaries, $I+2/3 = \dfrac{14-\sqrt{17}-\sqrt{21}+\sqrt{5}-\sqrt{13}}{2}$.
Correct Answer: (C) $\dfrac{14-\sqrt{17}-\sqrt{21}+\sqrt{5}-\sqrt{13}}{2}$