If int_0^x {f(t),dt} = x + int_x^1 {t,f(t),dt | vedclass

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(A) Given the equation: $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt}$ Rewrite the integral $\int_x^1 {t\,f(t)\,dt}$ as $-\int_1^x {t\,f(t)\,dt}$:

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$1/2$

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(A) Given the equation: $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt}$ Rewrite the integral $\int_x^1 {t\,f(t)\,dt}$ as $-\int_1^x {t\,f(t)\,dt}$: $\int_0^x {f(t)\,dt} = x - \int_1^x {t\,f(t)\,dt}$ Differentiating both sides with respect to $x$ using the Leibniz integral rule: $\frac{d}{dx} \left( \int_0^x {f(t)\,dt} \right) = \frac{d}{dx} (x) - \frac{d}{dx} \left( \int_1^x {t\,f(t)\,dt} \right)$ Applying the Fundamental Theorem of Calculus: $f(x) = 1 - xf(x)$ Rearranging the terms to solve for $f(x)$: $f(x) + xf(x) = 1$ $f(x)(1 + x) = 1$ $f(x) = \frac{1}{1 + x}$ Now,substitute $x = 1$ to find $f(1)$: $f(1) = \frac{1}{1 + 1} = \frac{1}{2}$.

If $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt,}$ then the value of $f(1)$ is

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