Question

​calcular essas integrais definidas.

Answer 1

a)

$\int\limits_{0}^{1}~(x^4-3x^2+1)dx~=~\left(\frac{x^{4+1}}{4+1}-\frac{3x^{2+1}}{2+1}+1.x^{0+1}\right|_{0}^{1}\\\\\\\\\int\limits_{0}^{1}~(x^4-3x^2+1)dx~=~\left(\frac{x^{5}}{5}-\frac{3x^{3}}{3}+x^{1}\right|_{0}^{1}\\\\\\\\\int\limits_{0}^{1}~(x^4-3x^2+1)dx~=~\left(\frac{x^{5}}{5}-x^{3}+x\right|_{0}^{1}\\\\\\\\\int\limits_{0}^{1}~(x^4-3x^2+1)dx~=~\left(\frac{1^5}{5}-1^3+1\right)-\left(\frac{0^5}{5}-0^3+0\right)$

$\int\limits_{0}^{1}~(x^4-3x^2+1)dx~=~\left(\frac{1}{5}-1+1\right)\\\\\\\\\boxed{\int\limits_{0}^{1}~(x^4-3x^2+1)dx~=~\frac{1}{5}}$

b)

$-x~=~u\\dx~=~-1.du\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~\int\limits_{0}^{\infty}~e^{u}~.~-1.du\\\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~-\int\limits_{0}^{\infty}~e^{u}du\\\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~\left(-e^u\right|_{0}^{\infty}\\\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~\left(-e^{-x}\right|_{0}^{\infty}\\\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~\left(-e^{-\infty}\right)-\left(-e^{-0}\right)$

$\int\limits_{0}^{\infty}~e^{-x}dx~=~\left(-\frac{1}{e^{\infty}}\right)-\left(-e^{0}\right)\\\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~\left(\to0\right)-\left(-1\right)\\\\\\\\\int\limits_{0}^{\infty}~e^{-x}dx~=~0+1\\\\\\\\\boxed{\int\limits_{0}^{\infty}~e^{-x}dx~=~1}$

c)

$\int\limits_{1}^{2}~\frac{2}{x}dx~=~2.\int\limits_{1}^{2}~\frac{1}{x}dx\\\\\\\\\int\limits_{1}^{2}~\frac{2}{x}dx~=~2.\left(\,ln(x)\right|_{1}^2\\\\\\\\\int\limits_{1}^{2}~\frac{2}{x}dx~=~2~.~(ln(2)-ln(1))\\\\\\\\\int\limits_{1}^{2}~\frac{2}{x}dx~=~2~.~(ln(2)-0)\\\\\\\\\boxed{\int\limits_{1}^{2}~\frac{2}{x}dx~=~2ln(2)}$

d)

$\int\limits_{0}^{\pi}~sin(x)dx~=~\left(-cos(x)\right|_{0}^{\pi}\\\\\\\\\int\limits_{0}^{\pi}~sin(x)dx~=~\left(-cos(\pi)\right)-\left(-cos(0)\right)\\\\\\\\\int\limits_{0}^{\pi}~sin(x)dx~=~\left(-(-1)\right)-\left(-(1)\right)\\\\\\\\\int\limits_{0}^{\pi}~sin(x)dx~=~1+1\\\\\\\\\boxed{\int\limits_{0}^{\pi}~sin(x)dx~=~2}$