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Question

Find the most general antiderivative of the function. (Check your answer by differentiation.)

f(x)=e2f ( x ) = e ^ { 2 }

Solution

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f(x)=e2Find an antiderivative F(x)=f(x)dxF(x)=e2dxF(x)=e2dxApply dx=x+C, soF(x)=e2x+CCheck by differentiationf(x)=F(x)f(x)=ddx[e2x+C]Apply ddx[x]=1, sof(x)=e2(1)+0f(x)=e2\begin{gathered} f\left( x \right) = {e^2} \\ \textcolor{#4257b2}{{\text{Find an antiderivative }}F\left( x \right) = \int {f\left( x \right)} dx} \\ F\left( x \right) = \int {{e^2}} dx \\ F\left( x \right) = {e^2}\int {dx} \\ \textcolor{#4257b2}{{\text{Apply }}\int {dx = x + C,{\text{ so}}}} \\ F\left( x \right) = {e^2}x + C \\ \textcolor{#4257b2}{{\text{Check by differentiation}}} \\ f\left( x \right) = F'\left( x \right) \\ f\left( x \right) = \frac{d}{{dx}}\left[ {{e^2}x + C} \right] \\ \textcolor{#4257b2}{ {\text{Apply }}\frac{d}{{dx}}\left[ x \right] = 1,{\text{ so}}} \\ f\left( x \right) = {e^2}\left( 1 \right) + 0 \\ f\left( x \right) = {e^2} \\ \end{gathered}

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