Question

Me ajudem por favor nessa questão de cálculo 1

Answer 1

Deseja-se provar que:

$\displaystyle \mathsf{\int_a^b\,x^2\,dx=\frac{b^3-a^3}{3}.}$

De início, aplique a seguinte regra:

$\displaystyle\mathsf{\int n^n\,dx=\frac{x^{n+1}}{n+1}+C,\,n\neq-1,\,C\in\mathbb{R}}$

Daí a integral indefinida fica:

$\displaystyle \mathsf{\int\,x^2\,dx=\frac{x^3}{3}+C}$

Agora, aplicando o Teorema Fundamental do Cálculo, segue que:

$\displaystyle \mathsf{\int_a^b\,x^2\,dx=\left.\frac{x^3}{3}\right|_a^b=}\\\\\\\mathsf{=\frac{b^3}{3}-\frac{a^3}{3}=}\\\\\\\mathsf{=\frac{b^3-a^3}{3}}$

Como queríamos demonstrar.

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Answer 2

$\displaystyle \underline{\text{Teorema Fundamental do C{\'a}lculo}}: \\\\ \int\limits^\text a_\text b \text{f(x)dx} = \text{F(a)} -\text{F(\text b)} \\\\\\ \text{Temos}: \\\\ \text{Prove que } \\\\ \int\limits^\text a_\text{b} \text{x} ^2\text{dx}=\frac{\text b^3-\text a^3}{3} \\\\\\ \underline{\text{Vamos mexer apenas no lado esquerdo}}: \\\\ \left[\begin{array}{c}\displaystyle \frac{\text x^3}{3}\end{array}\right] \limits^{\text a}_\text b = \frac{\text b^3-\text a^3}{3}$

$\displaystyle \frac{\text b^3}{3}-\frac{\text a^3}{3}=\frac{\text b^3-\text a^3}{3} \\\\\\ \huge\boxed{\frac{\text b^3-\text a^3}{3}=\frac{\text b^3-\text a^3}{3}\ }\checkmark \ \text{(C.Q.D)}$