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lorasvet [3.4K]

2 years ago

14

Pls help me!!!! I'm stuck on how to solve this!!??

1 answer:

Luda [366]2 years ago

6 0

Let A represent the value of the car after each year.
A= initial value (P)×(1+percent increase(r)) ^time
A=P×(1+r)^t
A=18710×(1+(-12%))^8
A=18710×(1-12%)^8
A=18710×(1-0.12)^8
A=18710×(0.88)^8
A= 6728.7619591115
The best approximation is 6729
Therefore the value of the car will be about $6729 after 8 years
Your answer is B.

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I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

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$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

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Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

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