Question

$\int\limits^2_0 { x^{2} 3^x \, dx$
por favor calculos... integral por partes

Answer 1

Resolver primeiro a integral indefinida

$I=\int{x^{2}3^{x}\mathrm{\,d}x}\\ \\ \\ I=\int{u\mathrm{\,d}v}\\ \\ \\ \begin{array}{ll} u=x^{2}\;\;&\;\;\mathrm{\,d}v=3^{x}\mathrm{\,d}x\\ du=2x\mathrm{\,d}x\;\;&\;\;v=\dfrac{3^{x}}{\mathrm{\ell n\,}3} \end{array}\\ \\ \\ I=uv-\int{v\mathrm{\,d}u}\\ \\ I=\dfrac{x^{2}3^{x}}{\mathrm{\ell n\,}3}-\int{\dfrac{3^{x}}{\mathrm{\ell n\,}3}\cdot 2x\mathrm{\,d}x}\\ \\ I=\dfrac{x^{2}3^{x}}{\mathrm{\ell n\,}3}-\dfrac{2}{\mathrm{\ell n\,}3}\underbrace{\int{x\cdot 3^{x}\mathrm{\,d}x}}_{I_{2}}$


$I_{2}=\int{x\cdot 3^{x}\mathrm{\,d}x}\\ \\ I_{2}=\int{u_{2}\mathrm{\,d}v_{2}}\\ \\ \\ \begin{array}{ll} u_{2}=x\;\;&\;\;\mathrm{\,d}v_{2}=3^{x}\mathrm{\,d}x\\ du_{2}=\mathrm{\,d}x\;\;&\;\;v_{2}=\dfrac{3^{x}}{\mathrm{\ell n\,}3} \end{array}\\ \\ \\ I_{2}=u_{2}v_{2}-\int{v_{2}\mathrm{\,d}u_{2}}\\ \\ I_{2}=\dfrac{x\cdot 3^{x}}{\mathrm{\ell n\,}3}-\int{\dfrac{3^{x}}{\mathrm{\ell n\,}3}\mathrm{\,d}x}\\ \\ I_{2}=\dfrac{x\cdot 3^{x}}{\mathrm{\ell n\,}3}-\dfrac{1}{\mathrm{\ell n\,}3}\int{3^{x}\mathrm{\,d}x}\\ \\ I_{2}=\dfrac{x\cdot 3^{x}}{\mathrm{\ell n\,}3}-\dfrac{1}{\mathrm{\ell n\,}3}\cdot \dfrac{3^{x}}{\mathrm{\ell n\,}3}\\ \\ I_{2}=\dfrac{x\cdot 3^{x}}{\mathrm{\ell n\,}3}-\dfrac{3^{x}}{\mathrm{\ell n}^{2\,}3}$


Substituindo de volta na expressão da integral original, temos

$I=\dfrac{x^{2}3^{x}}{\mathrm{\ell n\,}3}-\dfrac{2}{\mathrm{\ell n\,}3}\cdot \left(\dfrac{x\cdot 3^{x}}{\mathrm{\ell n\,}3}-\dfrac{3^{x}}{\mathrm{\ell n}^{2\,}3} \right )+C\\ \\ I=\dfrac{x^{2}3^{x}}{\mathrm{\ell n\,}3}-\dfrac{2x\cdot 3^{x}}{\mathrm{\ell n}^{2\,}3}+\dfrac{2\cdot 3^{x}}{\mathrm{\ell n}^{3\,}3}+C\\ \\ I=\left(\dfrac{x^{2}}{\mathrm{\ell n\,}3}-\dfrac{2x}{\mathrm{\ell n}^{2\,}3}+\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 3^{x}+C\\ \\ \boxed{\int{x^{2}3^{x}\mathrm{\,d}x}=\left(\dfrac{x^{2}}{\mathrm{\ell n\,}3}-\dfrac{2x}{\mathrm{\ell n}^{2\,}3}+\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 3^{x}+C}$


Substituindo os limites de integração temos que

$\int_{0}^{2}{x^{2}3^{x}\mathrm{\,d}x}\\ \\ =\left.\left(\dfrac{x^{2}}{\mathrm{\ell n\,}3}-\dfrac{2x}{\mathrm{\ell n}^{2\,}3}+\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 3^{x}\right]_{0}^{2}\\ \\ =\left(\dfrac{2^{2}}{\mathrm{\ell n\,}3}-\dfrac{2 \cdot 2}{\mathrm{\ell n}^{2\,}3}+\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 3^{2}- \left(\dfrac{0^{2}}{\mathrm{\ell n\,}3}-\dfrac{2 \cdot 0}{\mathrm{\ell n}^{2\,}3}+\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 3^{0}\\ \\ =\left(\dfrac{4}{\mathrm{\ell n\,}3}-\dfrac{4}{\mathrm{\ell n}^{2\,}3}+\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 9- \left(\dfrac{2}{\mathrm{\ell n}^{3\,}3} \right )\cdot 1\\ \\ =\dfrac{36}{\mathrm{\ell n\,}3}-\dfrac{36}{\mathrm{\ell n}^{2\,}3}+\dfrac{18}{\mathrm{\ell n}^{3\,}3}-\dfrac{2}{\mathrm{\ell n}^{3\,}3}\\ \\ =\dfrac{36}{\mathrm{\ell n\,}3}-\dfrac{36}{\mathrm{\ell n}^{2\,}3}+\dfrac{18-2}{\mathrm{\ell n}^{3\,}3}$

$=\dfrac{36}{\mathrm{\ell n\,}3}-\dfrac{36}{\mathrm{\ell n}^{2\,}3}+\dfrac{16}{\mathrm{\ell n}^{3\,}3}\\ \\ \\ \boxed{\int_{0}^{2}{x^{2}3^{x}\mathrm{\,d}x}=\dfrac{36}{\mathrm{\ell n\,}3}-\dfrac{36}{\mathrm{\ell n}^{2\,}3}+\dfrac{16}{\mathrm{\ell n}^{3\,}3}}$