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Thread: The Fundamental Theorem of Calculus

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    The Fundamental Theorem of Calculus

    4 integral 9 2xsquare root 2
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    Is this it: $\displaystyle \int_4^9 2 \sqrt{2} \ dx$

    Factor out the constant: $\displaystyle 2\sqrt{2} \int_4^9 dx$

    Use the fact that: $\displaystyle \int x^{n}\ dx = \frac{x^{n+1}}{n} + C$

    Here imagine $\displaystyle n = 0$ : $\displaystyle \int_4^9 x^0 \ dx$
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    I'm sorry inside the square root is x
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    Like this? $\displaystyle \int_4^92\sqrt{2x} \ dx$

    Remember that square roots can be written as exponent (1/2). So factor out the constant to get: $\displaystyle 2\int_4^9 (2x)^{\frac{1}{2}} \ dx$

    Use the sub: $\displaystyle u = 2x \ \Rightarrow \ du = 2 dx \ \Leftrightarrow \ \frac{du}{2} = dx$
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