Question

let f(t) be an odd function. using properties of the definite integral plus simple substitution show that if f is continuous on [-a,a] for a positive number a, then integral ∫a-af(t)dt=0

Answer 1

\[\int\limits_{-a}^{a}f(t) dt\]
we know
\[f(-t)=-f(t)\]


assuming a>0

we have

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-a 0 a

btw -a and 0 we have -f(t)
btw a and 0 we have f(t)

\[\int\limits_{-a}^{0}f(t)dt+\int\limits_{0}^{a}f(t) dt\]

\[-\int\limits_{0}^{-a}f(t)dt+\int\limits_{0}^{a}f(t) dt\]

let u=-t => du=-dt

\[-\int\limits_{0}^{a}f(-u) (-du)+\int\limits_{0}^{a}f(t) dt\]

\[-\int\limits_{0}^{a}f(u) du+\int\limits_{0}^{a}f(t) dt\]
\[-(F(a)-F(0))+(F(a)-F(0))=0\]

Answer 2

omg thank you soo much