Author Topic: Mathematics Computation (Read 3700 times) Tweet Share

12070 · « **on:** August 13, 2017, 03:38:26 pm »

Hey just wondering if any could help me with this: 1.) C

RuiAce · « **Reply #1 on:** August 13, 2017, 03:54:34 pm »

My computer hard drive crashed and I'm still yet to reinstall MATLAB. So I had to use Desmos to approximate the answer to 1b).

Desmos told me $ g(x)\approx 5.60517$ and the second intercept is at approximately $x\approx 1.827$. So I'll be using that instead.

$\text{It would appear as though the region between the blue curve}\\ \text{and the red curve defines our cross section of the trough}$
$\text{Hence, to find the cross-sectional area, we're interested in the area integral}\\ A = \int_0^{1.827}\left(g(x)-f(x)\right)\,dx$
$\text{For simplicity, I will break the integral up into}\\ A = \int_0^{1.827} 5.60517 \, dx+ \int_0^{1.827} \ln (x+0.01)\,dx - \int_0^{1.827}e^x\,dx$
_____________________________________
$\int_0^{1.827}5.60517 = 1.827\times 5.60517 \approx 10.2406$
$-\int_0^{1.827}e^x =- e^{1.827}+1 \approx -5.2152$
$\text{The second integral is trickier as we will inevitably require integration by parts.}$
_____________________________________
$\text{To avoid dealing with weird fractions}\\ \text{Let }u=x+0.01\implies du=dx\\ \text{Then the second integral becomes }\int_{0.01}^{1.837}\ln u \,<br />du$
$\text{To evaluate this,}\\ \begin{align*}\int_{0.01}^{1.837} \ln u\, du&= u\ln u \bigg|_{0.01}^{1.837} - \int_{0.01}^{1.837}\frac{u}{u} du\\ &= 1.837\ln1.837-0.01\ln0.01 - (1.837-0.01)\end{align*}\\ \text{The rest should be clear}$
Subject to minor computational errors.

12070 · « **Reply #2 on:** August 13, 2017, 05:20:07 pm »

Quote from: RuiAce on August 13, 2017, 03:54:34 pm

My computer hard drive crashed and I'm still yet to reinstall MATLAB. So I had to use Desmos to approximate the answer to 1b).
Subject to minor computational errors.

Thanks heaps Rui

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Author Topic: Mathematics Computation (Read 3700 times) Tweet Share

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