Question

Calcula a Integral Definida =

Answer 1

Integral por partes:

$\int\limits{u} \, dv=uv-\int\limits {v} \, du\\\\\\\int\limits {x^2e^x} \, dx \\\\u=x^2\to du=2x\ dx\\\\dv=e^x\ dx\to v=e^x\\\\\\\int\limits {x^2e^x} \, dx=x^2e^x-\int\limits {2xe^x} \, dx$

Agora o msm processo pra resolver aquela segunda integral que apareceu:

$\int\limits{u} \, dv=uv-\int\limits {v} \, du\\\\\\\int\limits {2xe^x} \, dx \\\\u=2x\to du=2\ dx\\\\dv=e^x\ dx\to v=e^x\\\\\\\int\limits {2xe^x} \, dx=2xe^x-\int\limits {2e^x} \, dx=2xe^x-2e^x$

Logo:

$\int\limits {x^2e^x} \, dx=\\\\\\x^2e^x-\int\limits {2xe^x} \, dx=\\\\\\x^2e^x-(2xe^x-2e^x)=\\\\\\x^2e^x-2xe^x+2e^x$

Agora aplicamos o intervalo:

$[x^2e^x-2xe^x+2e^x]^1_0=\\\\\\((1)^2e^{(1)}-2(1)e^{(1)}+2e^{(1)})-((0)^2e^{(0)}-2(0)e^{(0)}+2e^{(0)})=\\\\\\(e-2e+2e)-(2)=\boxed{e-2}$

Answer 2

• O resultado da sua integral definida é e - 2.

Para resolver sua questão, devemos aplicar a fórmula da integração por partes. Dada por:

$\large\displaystyle\begin{aligned}\int u\cdot dv \Leftrightarrow u\cdot v - \int v\cdot du\end{aligned}$

• Logo, aplicando na sua questão, temos que:

$\large\displaystyle\begin{aligned}\int ^1_0 x^2e^x dx\Leftrightarrow x^2 \cdot e^x - \int e^x 2x\ dx\end{aligned}$

$\large\displaystyle\begin{aligned}\int ^1_0 x^2e^x dx\Leftrightarrow x^2 \cdot e^x - 2\cdot\underbrace{ \int e^x x\ dx}_{\sf integre\ de\ novo}\end{aligned}$

$\large\displaystyle\begin{aligned}\int_0^1 x^2e^x dx\Leftrightarrow x^2 \cdot e^x - 2\cdot\left[ \left( xe^x - \int e^x dx \right) \right]\end{aligned}$

$\large\displaystyle\begin{aligned}\int_0^1 x^2e^x dx\Leftrightarrow x^2 \cdot e^x - 2\cdot\left[ \left( xe^x - e^x\right) \right]\end{aligned}$

$\large\displaystyle\begin{aligned}\int_0^1 x^2e^x dx\Leftrightarrow x^2 \cdot e^x - 2 xe^x + 2e^x\end{aligned}$

$\large\displaystyle\begin{aligned} \left. \left(x^2 \cdot e^x - 2 xe^x + 2e^x \right)\right|^1_0 \Leftrightarrow \end{aligned}$

$\large\displaystyle\begin{aligned} \left(1^2 \cdot e^{1} - 2 e^1 + 2e^1 \right) - \left(0^2\cdot e^0 -2\cdot 0e^0 + 2e^0\right)\Leftrightarrow \end{aligned}$

$\large\displaystyle\begin{aligned} \left( 1\cdot e \right) - \left(0\cdot 1-0 + 2\right)\Leftrightarrow \end{aligned}$

$\large\displaystyle\begin{aligned} \left( e \right) - \left(2\right)\Leftrightarrow \end{aligned}$

$\large\displaystyle\text{ \begin{aligned}\int ^1_0 x^2e^x dx\Leftrightarrow \underline{\boxed{\boxed{\green{ e-2}}}}\end{aligned} }$

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Integrais definidas.

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