Ensino Superior

IFB-2007 - Dada f(x): [0,1] [tex3]\rightarrow [/tex3] R+ contínua e diferenciável e f(0) = 1 e f(1) = 4. O valor da integral [tex3]\int\limits_{0}^{1}3.\sqrt{f(x)}.f'(x)dx[/tex3] é:
a) 12
b) 10
c) 0
d) 9
e) 14

Seja [tex3]\mathsf{I \ = \ \int 3\cdot f'(x) \cdot \sqrt{f(x)} \ dx}[/tex3] . Resolvendo-a usando o produto de funções:

[tex3]\mathsf{\hookrightarrow u \ = \sqrt{f(x)} \ \therefore \ u' \ = \ \dfrac{f'(x) \ dx}{2 \cdot \sqrt{f(x)}}}[/tex3] (regra da cadeia)

[tex3]\mathsf{\hookrightarrow \ v' \ = \ 3\cdot f'(x) \ dx \ \therefore \ v \ = \ 3\cdot f(x)}[/tex3]

[tex3]\mathsf{I \ = \ u \cdot v \ - \ \int v \cdot u'}[/tex3]

[tex3]\mathsf{I \ = \ 3\cdot f(x) \cdot \sqrt{f(x)} \ - \ \int 3\cdot f(x) \ \cdot \dfrac{f'(x) \ dx}{2 \cdot \sqrt{f(x)}}}[/tex3]

[tex3]\mathsf{I \ = \ 3\cdot f(x) \cdot \sqrt{f(x)} \ - \ \dfrac{1}{2} \cdot \underbrace{\mathsf{\int 3\cdot f'(x) \cdot \sqrt{f(x)} \ dx}}_{= \ I}}[/tex3]

[tex3]\mathsf{\dfrac{\cancel{3} \cdot I}{2} \ = \ \cancel{3}\cdot f(x) \cdot \sqrt{f(x)} }[/tex3]

[tex3]\boxed{\mathsf{I \ = \ 2 \cdot f(x) \cdot \sqrt{f(x)}}}[/tex3]

Logo, [tex3]\mathsf{\int\limits_{0}^{1} 3\cdot f'(x) \cdot \sqrt{f(x)} \ dx \ = \ 2 \cdot f(x) \cdot \sqrt{f(x)} \Bigg\vert_{0}^{1}}[/tex3]

[tex3]\mathsf{= \ 2 \cdot 4 \cdot 2 \ - \ 2 \cdot 1 \cdot 1}[/tex3]

[tex3]= \ \boxed{\boxed{\mathsf{14}}}[/tex3]

Ensino Superior ⇒ integral Tópico resolvido

integral

Re: integral

Ensino Superior ⇒ integral Tópico resolvido

integral

Re: integral

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