Ensino Superior ⇒ Cálculo 1 - Integral Tópico resolvido

[neonexe]

Seja f uma função com derivada contínua no intervalo [-1,1] tal que

f(-1)=[tex3]-4e^{3}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]-4e^{3}[/tex3]

e f(1)=[tex3]\frac{4}{e^{5}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{4}{e^{5}}[/tex3]

Sabendo que

[tex3]\int\limits_{-3}^{5}e^{u}f(\frac{u-1}{4})du[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{-3}^{5}e^{u}f(\frac{u-1}{4})du[/tex3]

= 4

determine o valor de

[tex3]\int\limits_{-1}^{1}e^{4x+1}f'(x)dx[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{-1}^{1}e^{4x+1}f'(x)dx[/tex3]

Resposta

---

4

[rcompany]

[tex3]
\begin{align}
\int_{-1}^{1}e^{4x+1}f'(x)\mathrm{d}x&=\left[e^{4x+1}f(x)\right]_{\!-\!1}^{\ 1}-\int_{-1}^1(e^{4x+1})'f(x)\mathrm{d}x\quad\small(f(x)g(x))'=f'(x)g(x)+f(x)g'(x))\\
&=e^5\cdot\dfrac{4}{e^5}+4e^{-3}e^3-\int_{-1}^14e^{4x+1}f(x)\mathrm{d}x\\
&=8-\int_{-3}^5 e^uf(\dfrac{u-1}{4})\mathrm{d}u\quad\text{com }u=4x+1,\,\text{ie. }x=\dfrac{u-1}{4}\text{ e }\mathrm{d}x=\dfrac{\mathrm{d}u}{4}\\
&=8-4=4

\end{align}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\begin{align}
\int_{-1}^{1}e^{4x+1}f'(x)\mathrm{d}x&=\left[e^{4x+1}f(x)\right]_{\!-\!1}^{\ 1}-\int_{-1}^1(e^{4x+1})'f(x)\mathrm{d}x\quad\small(f(x)g(x))'=f'(x)g(x)+f(x)g'(x))\\
&=e^5\cdot\dfrac{4}{e^5}+4e^{-3}e^3-\int_{-1}^14e^{4x+1}f(x)\mathrm{d}x\\
&=8-\int_{-3}^5 e^uf(\dfrac{u-1}{4})\mathrm{d}u\quad\text{com }u=4x+1,\,\text{ie. }x=\dfrac{u-1}{4}\text{ e }\mathrm{d}x=\dfrac{\mathrm{d}u}{4}\\
&=8-4=4

\end{align}
[/tex3]