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Definite Integration question

2021 · 16 Mar · Shift 1 · Q38
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Definite Integration question

2021 · 16 Mar · Shift 1 · Q38

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let f : R →\to→ R be a continuous function such that f(x) + f(x + 1) = 2, for all x ∈\in∈ R. If I1=∫08f(x)dx{I_1} = \int\limits_0^8 {f(x)dx}I1​=0∫8​f(x)dx and I2=∫−13f(x)dx{I_2} = \int\limits_{ - 1}^3 {f(x)dx}I2​=−1∫3​f(x)dx, then the value of I1 + 2I2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 16

  1. We are given f(x)+f(x+1)=2for all x∈R.f(x)+f(x+1)=2 \quad \text{for all } x\in\mathbb R.f(x)+f(x+1)=2for all x∈R.

  2. First, compute I1=∫08f(x) dx.I_1=\int_0^8 f(x)\,dx.I1​=∫08​f(x)dx.

    Split the interval into pairs of length 111: I1=∫01f(x)dx+∫12f(x)dx+⋯+∫78f(x)dx.I_1=\int_0^1 f(x)dx+\int_1^2 f(x)dx+\cdots+\int_7^8 f(x)dx.I1​=∫01​f(x)dx+∫12​f(x)dx+⋯+∫78​f(x)dx.

    Group them as: ∫01(f(x)+f(x+1))dx+∫23(f(x)+f(x+1))dx+∫45(f(x)+f(x+1))dx+∫67(f(x)+f(x+1))dx.\int_0^1 \bigl(f(x)+f(x+1)\bigr)dx + \int_2^3 \bigl(f(x)+f(x+1)\bigr)dx + \int_4^5 \bigl(f(x)+f(x+1)\bigr)dx + \int_6^7 \bigl(f(x)+f(x+1)\bigr)dx.∫01​(f(x)+f(x+1))dx+∫23​(f(x)+f(x+1))dx+∫45​(f(x)+f(x+1))dx+∫67​(f(x)+f(x+1))dx.

    Since f(x)+f(x+1)=2f(x)+f(x+1)=2f(x)+f(x+1)=2, I1=∫012 dx+∫232 dx+∫452 dx+∫672 dx=2+2+2+2=8.I_1=\int_0^1 2\,dx+\int_2^3 2\,dx+\int_4^5 2\,dx+\int_6^7 2\,dx = 2+2+2+2=8.I1​=∫01​2dx+∫23​2dx+∫45​2dx+∫67​2dx=2+2+2+2=8.

  3. Now compute I2=∫−13f(x) dx.I_2=\int_{-1}^3 f(x)\,dx.I2​=∫−13​f(x)dx.

    Again split into pairs: I2=∫−10f(x)dx+∫01f(x)dx+∫12f(x)dx+∫23f(x)dx.I_2=\int_{-1}^0 f(x)dx+\int_0^1 f(x)dx+\int_1^2 f(x)dx+\int_2^3 f(x)dx.I2​=∫−10​f(x)dx+∫01​f(x)dx+∫12​f(x)dx+∫23​f(x)dx.

    Group as: I2=∫−10(f(x)+f(x+1))dx+∫12(f(x)+f(x+1))dx.I_2=\int_{-1}^0 \bigl(f(x)+f(x+1)\bigr)dx + \int_1^2 \bigl(f(x)+f(x+1)\bigr)dx.I2​=∫−10​(f(x)+f(x+1))dx+∫12​(f(x)+f(x+1))dx.

    Therefore, I2=∫−102 dx+∫122 dx=2+2=4.I_2=\int_{-1}^0 2\,dx+\int_1^2 2\,dx=2+2=4.I2​=∫−10​2dx+∫12​2dx=2+2=4.

  4. Hence, I1+2I2=8+2(4)=8+8=16.I_1+2I_2=8+2(4)=8+8=16.I1​+2I2​=8+2(4)=8+8=16.

  5. Final answer: 16\boxed{16}16​

  6. Comparison with stored correct answer:

    Stored correct answer = 161616.

    Our derived answer matches it.

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