Question

Utilizando o Teorema Fundamental do Cálculo, analise as integrais definidas e determine cada um delas:

Answer 1

a)

de 1 a 9 ∫ 1/√x +x^(1/4)  dx

de 1 a 9 ∫ x^(-1/2) +x^(1/4+1) /(1/4+1) dx

de 1 a 9[ (x^(1/2) /(1/2] +x^(5/4)/(5/4] dx

de 1 a 9[ 2*(x^(1/2)  +(4/5)*x^(5/4)] dx

=[ 2*(9^(1/2)  +(4/5)*9^(5/4)] - [ 2*(1^(1/2)  +(4/5)*1^(5/4)]

=[ 2*3 +(4/5)*3^(5/2)] - [ 2  +4/5]

~  15,670

b)

0 a 2  ∫√(2x) * (√x+√5) dx

0 a 2  ∫x√(2)  + √(10x) dx

0 a 2 [ x²√(2)/2 +√(10)*(x)^(1/2+1)/(1/2+1)]

0 a 2 [ x²√(2)/2 +(2/3)*√(10) (x)^(3/2)]

=2²√(2)/2 +(2/3)*√(10) (2)^(3/2)

=2√2 +(2/3)* √(10) * (2)^(3/2) ~  8.7912750

c)

1 a 32 ∫ (1+x^(2/5))/x^(1/3) dx

1 a 32 ∫ x^(-1/3) +x^(2/5-1/3 )  dx

1 a 32 ∫ x^(-1/3) +x^[(6-5 )/15]  dx

1 a 32 ∫ x^(-1/3) +x^(1 /15]  dx

1 a 32 [ x^(-1/3+1) /(-1/3+1)+ x^(16/15)/(16/15)]

1 a 32 [ (3/2)*x^(2/3) + (15/16)*x^(16/15)]

=[ (3/2)*32^(2/3) +  (15/16)*32^(16/15)] -[ (3/2)*1^(2/3) +  (15/16)*1^(16/15)]

~ 50,4791

Answer 2

                                          Integral <<<

• A)

$\sf Regra\Rightarrow~\boxed{\sf \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}$

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$\sf \displaystyle \large{\sf \int _1^9\frac{1}{\sqrt{x}}+\sqrt[\sf 4]{\sf x}~dx}\\\\\\\\\sf \int _1^9\frac{1}{\sqrt{x}}dx+\int _1^9\sqrt[\sf 4]{\sf x}~dx\\\\\\\\\sf \int _1^9\frac{1}{\sqrt{x}}dx=4\\\\\\\\\sf \int _1^9\sqrt[\sf 4]{\sf x}~dx=4+\frac{36\sqrt{3}}{5}-\frac{4}{5}\\\\\\\\\sf 4+\frac{36\sqrt{3}}{5}-\frac{4}{5}\\\\\\\\\boxed{\boxed{\boxed{\sf \frac{16}{5}+\frac{36\sqrt{3}}{5}\approx15,67076\cdots}}}$

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• B)

$\sf \displaystyle \int _0^2\:\sqrt{2x}\left(\sqrt{x}+\sqrt{5}\right)dx\\\\\\\\\sf \sqrt{2x}=\sqrt{2}~ \sqrt{x} \\\\\\\sf \int _0^2\sqrt{2}\sqrt{x}\left(\sqrt{x}+\sqrt{5}\right)dx\\\\\\\\\sf{Remova\:a\:constante}\Rightarrow\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\\\\\\sf \sqrt{2}\cdot \int _0^2\sqrt{x}\left(\sqrt{x}+\sqrt{5}\right)dx\\\\\\\\\sf \sqrt{2}\cdot \int _0^2x+\sqrt{5}\sqrt{x}dx$

$\sf \displaystyle {Aplique\:a\:regra\:da\:soma}\Rightarrow\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\\\\\sf \sqrt{2}\left(\int _0^2xdx+\int _0^2\sqrt{5}\sqrt{x}dx\right)\\\\\\\\\sf \boxed{\boxed{\boxed{\sf \sqrt{2}\left(2+\frac{4\sqrt{10}}{3}\right)\approx 8,79127\cdots}}}$

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• C)

$\sf \displaystyle \large{\int _1^{32}\:\frac{1+\sqrt[\sf 5]{\sf x^2}}{\sqrt[\sf 3]{\sf x}}\:dx}\\\\\\\\\sf \large{ \sqrt[\sf 5]{\sf x^2}=x^{\frac{2}{5}},\:\quad {\:assumindo\:que}\:x\ge 0}\\\\\\\\\sf \int _1^{32}\frac{1+x^{\frac{2}{5}}}{\sqrt[\sf 3]{\sf x}}dx\\\\\\\\\sf \int _1^{32}\frac{1}{\sqrt[\sf 3]{\sf x}}+\sqrt[\sf 15]{\sf x}dx\\\\\\\\ {Aplique\:a\:regra\:da\:soma}\Rightarrow\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$\sf \displaystyle \large{\int _1^{32}\frac{1}{\sqrt[\sf 3]{\sf x}}~dx+\int _1^{32}\sqrt[\sf 15]{\sf x}~dx}\\\\\\\\\sf 12\sqrt[\sf 3]{\sf 2}-\frac{3}{2}+30\sqrt[\sf 3]{\sf 2}-\frac{15}{16}\\\\\\\\\boxed{\boxed{\boxed{\sf 42\sqrt[\sf 3]{\sf 2}-\frac{39}{16}\approx50,47918\cdots}}}$

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