Definite IntegrationhardFree

Definite Integration: Continuous Differentiable Function

JEE Maths question with a full step-by-step solution.

Question
If ff is a continuous and differentiable function in x∈(0,1)x \in \left(0,1\right) such that
∑r=010(f(x+r)−∣ex−r−1∣)=0,\sum_{r=0}^{10}\left(f\left(x+r\right)-\left|e^x-r-1\right|\right) = 0 ,
 then ∫011f(x)dx\displaystyle\int_0^{11}f\left(x\right)dx is, where e=2.78e = 2.78, ln⁡2=0.693\ln2 = 0.693 and ln⁡3=1.098\ln3 = 1.098.
Solution
Answer: 48.31 (± 0.01)
Step 1: The condition is one functional identity:
f(x)+f(x+1)+⋯+f(x+10)=∣ex−1∣+∣ex−2∣+⋯+∣ex−11∣,x∈(0,1)f\left(x\right)+f\left(x+1\right)+\cdots+f\left(x+10\right) = \left|e^x-1\right|+\left|e^x-2\right|+\cdots+\left|e^x-11\right| ,\qquad x \in \left(0,1\right)
 Step 2: Substituting x↦x+rx \mapsto x+r in each unit piece,
∫011f(x)dx=∑r=010∫rr+1f(x)dx=∑r=010∫01f(x+r)dx=∫01[∑r=010f(x+r)]dx\int_0^{11}f\left(x\right)dx = \sum_{r=0}^{10}\int_r^{r+1}f\left(x\right)dx = \sum_{r=0}^{10}\int_0^{1}f\left(x+r\right)dx = \int_0^1\left[\sum_{r=0}^{10}f\left(x+r\right)\right]dx
∫011f(x)dx=∫01∑k=111∣ex−k∣ dx\int_0^{11}f\left(x\right)dx = \int_0^1\sum_{k=1}^{11}\left|e^x-k\right|\,dx
 Step 3: On (0,1)\left(0,1\right), 1<ex<e<31<e^x<e<3, so
∣ex−1∣=ex−1,∣ex−k∣=k−ex for k≥3 (since ex<e<3)\left|e^x-1\right| = e^x-1 ,\qquad \left|e^x-k\right| = k-e^x \ \text{for } k \ge 3 \ \left(\text{since } e^x<e<3\right)
 while ∣ex−2∣\left|e^x-2\right| changes sign at x=ln⁡2x = \ln2. Step 4: Using 3+4+⋯+11=633+4+\cdots+11 = 63, the terms without a modulus give
(ex−1)+∑k=311(k−ex)=ex−1+(3+4+⋯+11)−9ex=−8ex+62\left(e^x-1\right)+\sum_{k=3}^{11}\left(k-e^x\right) = e^x-1+\left(3+4+\cdots+11\right)-9e^x = -8e^x+62
∫01(−8ex+62)dx=−8(e−1)+62=−8e+70\int_0^1\left(-8e^x+62\right)dx = -8\left(e-1\right)+62 = -8e+70
 Step 5: Splitting the modulus term at x=ln⁡2x = \ln2,
∫0ln⁡2(2−ex)dx+∫ln⁡21(ex−2)dx=[2x−ex]0ln⁡2+[ex−2x]ln⁡21\int_0^{\ln2}\left(2-e^x\right)dx+\int_{\ln2}^{1}\left(e^x-2\right)dx = \left[2x-e^x\right]_0^{\ln2}+\left[e^x-2x\right]_{\ln2}^{1}
=(2ln⁡2−2+1)+(e−2−2+2ln⁡2)=4ln⁡2+e−5= \left(2\ln2-2+1\right)+\left(e-2-2+2\ln2\right) = 4\ln2+e-5
 Step 6:
∫011f(x)dx=(−8e+70)+(4ln⁡2+e−5)=65+4ln⁡2−7e\int_0^{11}f\left(x\right)dx = \left(-8e+70\right)+\left(4\ln2+e-5\right) = 65+4\ln2-7e
 With e=2.78e = 2.78 and ln⁡2=0.693\ln2 = 0.693 as given,
65+4(0.693)−7(2.78)=65+2.772−19.46=48.312  ⟶  48.3165+4\left(0.693\right)-7\left(2.78\right) = 65+2.772-19.46 = 48.312 \;\longrightarrow\; 48.31
 Answer: 65+4ln⁡2−7e≈48.3165+4\ln2-7e \approx 48.31
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