Let f(x) = int_{0}^{x} g(t) dt,where g is a n | vedclass

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(A) Given $f(x) = \int_{0}^{x} g(t) dt$. Since $g(t)$ is an even function,$f(x)$ is an odd function because $f(-x) = \int_{0}^{-x} g(t) dt$. Let $t =

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$\int_{x+5}^{5} g(t) dt$

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(A) Given $f(x) = \int_{0}^{x} g(t) dt$. Since $g(t)$ is an even function,$f(x)$ is an odd function because $f(-x) = \int_{0}^{-x} g(t) dt$. Let $t = -u$,then $dt = -du$. So $f(-x) = \int_{0}^{x} g(-u) (-du) = -\int_{0}^{x} g(u) du = -f(x)$.Given $f(x+5) = g(x)$. Since $g(x)$ is even,$g(x) = g(-x)$,so $f(x+5) = g(-x)$.Also,since $f$ is odd,$f(x+5) = -f(-x-5)$.We need to evaluate $I = \int_{0}^{x} f(t) dt$.From $f(x+5) = g(x)$,we have $g(t) = f(t+5)$.Since $g$ is even,$g(t) = g(-t)$,thus $f(t+5) = f(-t+5)$.Using the derivative,$f'(x) = g(x)$.Integrating $f(x+5) = g(x)$ from $0$ to $x$:$\int_{0}^{x} f(t+5) dt = \int_{0}^{x} g(t) dt = f(x)$.Let $u = t+5$,then $du = dt$. When $t=0, u=5$; when $t=x, u=x+5$.So,$\int_{5}^{x+5} f(u) du = f(x)$.This does not directly give the integral of $f$. Let's use $f'(x) = g(x)$.Since $f(x+5) = g(x)$,then $f'(x+5) = g'(x)$.Integrating $f(x+5) = g(x)$ with respect to $x$:$\int_{0}^{x} f(t+5) dt = \int_{0}^{x} g(t) dt = f(x)$.Using the property $\int_{0}^{x} f(t) dt = -\int_{0}^{x} f(-t) dt$,and $f$ is odd,$f(-t) = -f(t)$.Actually,from $f(x+5) = g(x)$,we have $f(x) = \int_{0}^{x} g(t) dt = \int_{0}^{x} f(t+5) dt$.Let $u = t+5$,then $\int_{5}^{x+5} f(u) du = f(x)$.Thus $\int_{0}^{x} f(t) dt = \int_{x+5}^{5} g(t) dt$ is the correct transformation.

Let $f(x) = \int_{0}^{x} g(t) dt$,where $g$ is a non-zero even function. If $f(x+5) = g(x)$,then $\int_{0}^{x} f(t) dt$ equals

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