Name: Iteration Method
Uploaded: 2019-07-19
Duration: 12 min 36 s
Description: In this video, we will learn iterative methods to solve systems of linear equations through the algorithm presented in this video. Watch and learn.

Question

It is given that $\int_{1}^{a} \ln x d x = 5,$ where $a$ is a constant greater than $1 .$

Use an iterative formula based on the equation in to find the value of $a$ correct to $3$ decimal places. Use an initial value of $5$ and give the result of each iteration to $5$ decimal places.

Data: $\ln (5) = 1.6094$ , $\ln (5.59201) \approx 1.73906, \ln (5.57241) \approx 1.71782 and \ln (5.57239) \approx 1.71782$

Accepted Answer

Given that $\int_{1}^{a} \ln x d x = 5$

Integration by parts requires a product of two terms in the integrand.

Using integrating by parts,

$= \int_{1}^{a} 1 (\ln x) d x$

Using the formula, $\int f (x) g (x) d x = f (x) \int g (x) d x - \int [f^{'} (x) \int g (x) d x] d x$

$= {[\ln x \int_{} 1 d x]}_{1}^{a} - \int_{1}^{a} (\frac{d (\ln x)}{d x} \int_{} 1 d x) d x$

$= {[x \ln x]}_{1}^{a} - \int_{1}^{a} (x . \frac{1}{x}) d x$

$= {[x \ln x]}_{1}^{a} - \int_{1}^{a} 1 d x$

$= {[x \ln x]}_{1}^{a} - {[x]}_{1}^{a}$

$= a \ln a - 1 \ln (1) - (a - 1)$

$= a \ln a - (a - 1)$

Using the fact that $\int_{1}^{a} \ln x d x = 5,$ where $a$ is a constant greater than $1 .$

$\Rightarrow a \ln a - (a - 1) = 5$

$\Rightarrow a \ln a - a + 1 = 5$

$\Rightarrow a \ln a = 4 + a$

$\Rightarrow a = \frac{4 + a}{\ln a}$

The iterative formula to find the value of $a$ is $a_{n + 1} = \frac{4 + a_{n}}{\ln α_{n}}$

The starting value of $a_{1} = 5$

$a_{1} = 5$ Are these same correct to $3$ significant figures

$a_{2} = \frac{4 + 5}{\ln (5)} = \frac{9}{\ln (5)} = 5.59201$ No

$a_{3} = \frac{4 + 5.59201}{\ln (5.59201)} = 5.57241$ No

$a_{4} = \frac{4 + 5.57241}{\ln (5.57241)} = 5.57239$ No

$a_{5} = \frac{4 + 5.57239}{\ln (5.57239)} = 5.57239$ Yes

$∴ a \approx 5.572$ (rounded to $3$ decimal places)

Important Questions on Numerical Solutions of Equations

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