Integration question (1 Viewer)

Farhanthestudent005 · Jun 1, 2022

Will appreciate a solution to this question...thanks

ExtremelyBoredUser · Jun 1, 2022

Farhanthestudent005 said:
View attachment 35702

Will appreciate a solution to this question...thanks

$u = 1 - x \, \text{so} \,du = - dx$ and x = 1 - u

The new bounds will be $u_{upper} = 1 - 0 = 1$ and $u_{bottom} = 1 - (-3) = 4$

$\Rightarrow -\int_{4}^{1} \frac{1-u}{\sqrt{u}} \, du$
$\Rightarrow \int_{1}^{4} \frac{1-u}{\sqrt{u}} \, du$
$\Rightarrow \int_{1}^{4} \frac{1}{\sqrt{u}} - \sqrt{u} \, du$
$2\sqrt{u} -\frac{2}{3} \times u\sqrt{u} ]_{1}^{4}$

Rest is just evaluating

Drongoski · Jun 1, 2022

Again, if not required to use the substitution:

$\int ^0 _{-3} \frac {x}{\sqrt {1-x} }dx = -\int ^0 _{-3}\frac {1-x - 1}{\sqrt {1-x} }dx\\ \\ = \int ^0 _{-3} (1-x)^{0.5} d(1-x)-\int ^0 _{-3}(1-x)^{-0.5}d(1-x)\\ \\ = \left[ \frac{2}{3}(1-x)^{1.5} -2(1-x)^{0.5} \right]^0 _{-3} \\ \\ =-(\frac {2}{3} - 2) +(\frac {2}{3}\times 8 - 2\times 2)\\ \\ = -\frac {8}{3}$

?????

Note: dx = -d(1-x)
No need to change limits of integration.

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Integration question (1 Viewer)

Farhanthestudent005

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ExtremelyBoredUser

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Drongoski

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