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\[ \large g(x) = e^x \int_0^1 e^x g(x) \, dx \]

Find \(g(0) \).

Note by D K 1 year, 4 months ago

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\[\large g(x)=e^x \underbrace{\displaystyle\int_0^1e^xg(x)\mathrm{d}x}_{\text{Constant=C}}=Ce^x\]

\[C=\displaystyle\int_0^1e^x(Ce^x)\mathrm{d}x=\frac C2(e^2-1)\implies C=0\]

Hence \(\large g(x)=0\forall x\in \mathbb R\)

Hence \(g(0)=0\). – Rishabh Cool · 1 year, 4 months ago

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@Rishabh Cool – Explain, how is that a constant? – Rishabh Kumar · 1 year, 4 months ago

@Rishabh Kumar – Any definite integral with constant bounds is a constant. – Andrew Ellinor · 1 year, 4 months ago

Couldn't \(g(x) = 0?\) Just seems like \(g(0) = 0\) then. – Andrew Ellinor · 1 year, 4 months ago

g(0)=0 – Gaurav Chahar · 1 year, 4 months ago

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## Comments

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TopNewest\[\large g(x)=e^x \underbrace{\displaystyle\int_0^1e^xg(x)\mathrm{d}x}_{\text{Constant=C}}=Ce^x\]

\[C=\displaystyle\int_0^1e^x(Ce^x)\mathrm{d}x=\frac C2(e^2-1)\implies C=0\]

Hence \(\large g(x)=0\forall x\in \mathbb R\)

Hence \(g(0)=0\). – Rishabh Cool · 1 year, 4 months ago

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– Rishabh Kumar · 1 year, 4 months ago

Explain, how is that a constant?Log in to reply

– Andrew Ellinor · 1 year, 4 months ago

Any definite integral with constant bounds is a constant.Log in to reply

Couldn't \(g(x) = 0?\) Just seems like \(g(0) = 0\) then. – Andrew Ellinor · 1 year, 4 months ago

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g(0)=0 – Gaurav Chahar · 1 year, 4 months ago

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