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If $f (0) = 2, f^{'} (x) = f (x), ϕ (x) = x + f (x)$ then $\int_{0}^{1} f (x) ϕ (x) d x$ is

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Correct option is B

Since f′(x)=f(x) and f(0)=2 so f(x)=2ex, thus ϕ(x)=x+2ex. Hence ∫01 f(x)ϕ(x)dx=2∫01 xex+2e2xdx=2xex−ex01+e2x01=2([e−e]+1)+e2−1=2e2.

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If f(0)=2, f′(x)=f(x), ϕ(x)=x+f(x) then ∫01 f(x)ϕ(x)dx is

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If $f (0) = 2, f^{'} (x) = f (x), ϕ (x) = x + f (x)$ then $\int_{0}^{1} f (x) ϕ (x) d x$ is