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JEE MainMathematicsModerateDefinite Integration

If
∫0100\mspace2muf(x)dx=a, then  ∑r=1100\mspace2mu∫01\mspace2muf(r−1+x)dx)=\int_{0}^{100}\mspace{2mu} f(x)dx = a,\ then\ \left. \ \sum_{r = 1}^{100}\mspace{2mu}\int_{0}^{1}\mspace{2mu} f(r - 1 + x)dx \right) =

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∑r=1100\mspace2mu(∫01\mspace2muf(r−1+x)dx)=∫01\mspace2muf(x)dx+∫01\mspace2muf(1+x)dx+∫01\mspace2muf(2+x)dx\sum_{r = 1}^{100}\mspace{2mu}\left( \int_{0}^{1}\mspace{2mu} f(r - 1 + x)dx \right) = \int_{0}^{1}\mspace{2mu} f(x)dx + \int_{0}^{1}\mspace{2mu} f(1 + x)dx + \int_{0}^{1}\mspace{2mu} f(2 + x)dx
+......+ ∫01\mspace2mu\int_{0}^{1}\mspace{2mu}f (99 + x) dx
=∫01\mspace2muf(x)dx+∫12\mspace2muf(x)dx= \int_{0}^{1}\mspace{2mu} f(x)dx + \int_{1}^{2}\mspace{2mu} f(x)dx+.....+∫99100\mspace2muf(x)dx\int_{99}^{100}\mspace{2mu} f(x)dx
=∫0100\mspace2muf(x)dx=a= \int_{0}^{100}\mspace{2mu} f(x)dx = a

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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