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Evaluate: ='false' \int^{+\infty} _6 \frac {1}{^3 \sqrt {x - 5}} dx

Question:

Evaluate: {eq}\int^{+\infty} _6 \frac {1}{^3 \sqrt {x - 5}} dx {/eq}

Integration:

Integration is the area under the curve between the defined intervals and

the improper integral is shown as: {eq}\int f(x)dx {/eq}

Answer and Explanation:

{eq}\begin{aligned} \int \frac{1}{\sqrt[3]{x-5}} & \mathrm{d} x \\ \text { Substitute } u=x-5 & \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=1(\text { steps }) \longrightarrow \mathrm{d} x=\mathrm{d} u : \\ &=\int \frac{1}{\sqrt[3]{u}} \mathrm{d} u \end{aligned}\\ \begin{array}{c}{\text { Apply power rule: }} \\ {\int u^{\mathrm{n}} \mathrm{d} u=\frac{u^{\mathrm{n}+1}}{\mathrm{n}+1} \text { with } \mathrm{n}=-\frac{1}{3}} \\ {=\frac{3 u^{\frac{2}{3}}}{2}} \\ {\text { Undo substitution } u=x-5}\end{array}\\ =\frac{3(x-5)^{\frac{2}{3}}}{2}+c {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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