If k int_{0}^{1} x cdot f(3x) , dx = int_{0}^ | vedclass

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(A) Let $I = k \int_{0}^{1} x \cdot f(3x) \, dx$. Substitute $t = 3x$,then $dt = 3 \, dx$,which implies $dx = \frac{dt}{3}$. When $x = 0$,$t = 0$. Whe

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$9$

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(A) Let $I = k \int_{0}^{1} x \cdot f(3x) \, dx$. Substitute $t = 3x$,then $dt = 3 \, dx$,which implies $dx = \frac{dt}{3}$. When $x = 0$,$t = 0$. When $x = 1$,$t = 3$. Substituting these into the integral: $I = k \int_{0}^{3} \left(\frac{t}{3}\right) \cdot f(t) \cdot \left(\frac{dt}{3}\right)$ $I = \frac{k}{9} \int_{0}^{3} t \cdot f(t) \, dt$. Given that $I = \int_{0}^{3} t \cdot f(t) \, dt$,we equate the two expressions: $\frac{k}{9} \int_{0}^{3} t \cdot f(t) \, dt = \int_{0}^{3} t \cdot f(t) \, dt$. Comparing the coefficients,we get $\frac{k}{9} = 1$,which implies $k = 9$.

If $k \int_{0}^{1} x \cdot f(3x) \, dx = \int_{0}^{3} t \cdot f(t) \, dt$,then the value of $k$ is

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