Question
If $\frac{d y}{d x}=x+\int_{0}^{1} y(x) d x, y(0)=1$, then, $\int_{0}^{1} x y(x) d x$ is(a) $\frac{13}{36}$(b) $\frac{19}{48}$(c) $\frac{7}{36}$(d) $\frac{101}{72}$
Step 1
Let's assume $\int_{0}^{1} y(x) d x = k$ for simplicity. So, the differential equation becomes $\frac{d y}{d x}=x+k$. Show more…
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