Question:
Let \(f:[-2,3]\to[0,\infty)\) be a continuous function such that \(f(1-x)=f(x)\) for all \(x\in[-2,3]\). If \(R_1\) is the numerical value of the area of the region bounded by \(y=f(x),\ x=-2,\ x=3\) and the axis of \(x\) and \(R_2=\int_{-2}^{3}xf(x)\,dx\), then :
Let \(f:[-2,3]\to[0,\infty)\) be a continuous function such that \(f(1-x)=f(x)\) for all \(x\in[-2,3]\). If \(R_1\) is the numerical value of the area of the region bounded by \(y=f(x),\ x=-2,\ x=3\) and the axis of \(x\) and \(R_2=\int_{-2}^{3}xf(x)\,dx\), then :
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Use \(\int_a^b g(x)dx=\int_a^b g(a+b-x)dx\) with \(a+b=1\).
Updated On: Nov 4, 2026
- \(3R_1=2R_2\)
- \(2R_1=3R_2\)
- \(R_1=R_2\)
- \(R_1=2R_2\)
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The Correct Option is D
Solution and Explanation
Since \(f\ge 0\), \(R_1=\int_{-2}^{3}f(x)\,dx\).
Here \(a+b=-2+3=1\), so replace \(x\) by \(1-x\): \(R_2=\int_{-2}^{3}(1-x)f(1-x)\,dx=\int_{-2}^{3}(1-x)f(x)\,dx\) (using \(f(1-x)=f(x)\)).
So \(R_2=\int f\,dx-\int xf\,dx=R_1-R_2\).
\(2R_2=R_1\), i.e. \(R_1=2R_2\).
Answer: (4) \(R_1=2R_2\)
Here \(a+b=-2+3=1\), so replace \(x\) by \(1-x\): \(R_2=\int_{-2}^{3}(1-x)f(1-x)\,dx=\int_{-2}^{3}(1-x)f(x)\,dx\) (using \(f(1-x)=f(x)\)).
So \(R_2=\int f\,dx-\int xf\,dx=R_1-R_2\).
\(2R_2=R_1\), i.e. \(R_1=2R_2\).
Answer: (4) \(R_1=2R_2\)
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