Evaluate the integral. ?

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Evaluate the integral. ?

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Answer 1 · 2017-12-14T03:51:05

3

Explanation:

Using u-substitution and the rules of anti derivatives.
Let $u = 27 + 2 x$ $u=27+2x$
$d u = 2 d x$ $du=2dx$ ; $\frac{d u}{2} = d x$ $(du)/2=dx$ or $d u \cdot \frac{1}{2} = d x$ $du*1/2=dx$
Go back to the original formula and plug that in
$\int_{0}^{49} (\frac{1}{\sqrt[3]{u^{2}}} d u \cdot \frac{1}{2})$ $int_0^49(1/root(3)(u^2)du*1/2)$
Using properties of integrals, I can put constants in front of the integral.
$\frac{1}{2} \int_{0}^{49} (\frac{1}{\sqrt[3]{u^{2}}}) d u$ $1/2int_0^49(1/root(3)(u^2))du$
Using properties of roots and exponents:
$a^{\frac{x}{y}} = \sqrt[y]{a^{x}}$ $a^(x/y)=root(y)(a^x)$
and $\frac{1}{x} = x^{- 1}$ $1/x=x^-1$
Now convert again to
$\frac{1}{2} \int_{0}^{49} (u^{- \frac{2}{3}}) d u$ $1/2int_0^49(u^(-2/3))du$
Apply reverse power rule to u and then get
$\frac{1}{2} {[3 \sqrt[3]{u}]}_{0}^{49}$ $1/2[3root(3)(u)]_0^49$
Replace u with original function and remember
$\int_{a}^{b} f (x) d x = F (b) - F (a)$ $int_a^bf(x)dx=F(b)-F(a)$
so then
$\frac{1}{2} [3 \sqrt[3]{27 + 2 (49)} - 3 \sqrt[3]{27 + 2 (0)}]$ $1/2[3root(3)(27+2(49))-3root(3)(27+2(0))]$ and you should hopefully get 3.
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Evaluate the integral. ?

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