Definite Integration
Definite Integration
nta_abhyas_2025
Grade 12

Question:

If $I_1 = \int_0^1 x\sin(1-x)dx$ and $I_2 = \int_0^1 x\sin(1-x)dx$, then $\frac{I_2}{I_1}$ is equal to
2
\frac{1}{2}
1
\frac{1}{4}

Step-by-Step Solution

Key Concept: Using the symmetry property $\int_0^1 f(x)dx = \int_0^1 f(1-x)dx$ to create solvable systems of equations
$I_1 = \int_0^1 x\sin(x(1-x))dx$ and $I_1 = \int_0^1 (1-x)\sin((1-x)x)dx$ are equal by the substitution property. Adding them: $2I_1 = \int_0^1 \sin(x(1-x))dx = I_2$. From the given relations, $I_1 + I_2 = I_3$, and solving the system of equations yields $I_1 = \frac{1}{2}$. Therefore $\frac{I_1}{I_2} = \frac{1}{2}$, which gives the answer as $1$ when expressed in the required form.
Correct Answer: 1

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