Question

integrais definida de ∫ 6x-1 dx ( limites 0 e 2)

Answer 1

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Integral da potência

$\large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int x^n~dx=\dfrac{x^{n+1}}{n+1}+k}}}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int_0^2(6x-1)~dx=6\int_0^2 x~dx-\int_0^2~dx\\\displaystyle\sf\int_0^2(6x-1)~dx=\bigg[ 6\cdot\dfrac{x^{1+1}}{1+1}-x\bigg]_0^2\\\displaystyle\sf\int_0^2(6x-1)~dx=\bigg[6\cdot\dfrac{x^2}{2}-x\bigg]_0^2\\\displaystyle\sf\int_0^2(6x-1)~dx=\bigg[3x^2-x\bigg]_0^2\\\sf como~o~integrando~cont\acute em~parte~nula,basta~fazer\\\sf a~substituic_{\!\!,}\tilde ao~do~limite~superior.\\\displaystyle\sf\int_0^2(6x-1)~dx=3\cdot2^2-2\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int_0^2(6x-1)~dx=12-2\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int_0^2(6x-1)~dx=10}}}}\end{array}}\blue{\checkmark}$

Answer 2

Olá,

Temos a integral:

$\tt \int \limits^{2}_{0}(6x - 1)dx$

Podemos resolver aplicando as propriedades elementares das integrais:

$\tt \int \limits^{2}_{0}(6x - 1)dx = \tt \int \limits^{2}_{0}6x dx - \tt \int \limits^{2}_{0}dx \\ \tt \int \limits^{2}_{0}(6x - 1)dx = 6 \frac{ {x}^{2} }{2} | {}^{2} _{0} - x {|}^{2}_{0} \\ \tt \int \limits^{2}_{0}(6x - 1)dx = \ 3{x}^{2} | {}^{2} _{0} - x {|}^{2}_{0} \\ \tt \int \limits^{2}_{0}(6x - 1)dx = 3( {2}^{2} - {0}^{2} ) - (2 - 0) \\ \tt \int \limits^{2}_{0}(6x - 1)dx = 3(4 - 0) - 2 \\ \tt \int \limits^{2}_{0}(6x - 1)dx = 12 - 2 \\ \boxed{\tt \int \limits^{2}_{0}(6x - 1)dx = 10}$