گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!

a)  Given that $\int_1^\alpha  {\frac{{\ln x}}{{{x^2}}}dx}  = \frac{2}{5}$,  show that $\alpha  = \frac{6}{3}\left( {1 + \ln \alpha } \right)$.

b) Use an iteration formula based on the equatio $\alpha  = \frac{5}{3}\left( {1 + \ln \alpha } \right)$  to find the value of $\alpha $  correct to 2 decimal places. Use an initial value of 4 and give the result of each iteration to 4 decimal places.

پاسخ تشریحی :
نمایش پاسخ

a) Attempt integration by parts

Obtain $ - {x^{ - 1}}\ln x + \int {\frac{1}{{{x^2}}}dx} ,\,\,\frac{{x\ln x - x}}{{{x^2}}} + 2\int {\frac{{\ln x}}{{{x^2}}}dx - 2\int {\frac{1}{{{x^2}}}dx} } $

Obtain $ - {x^{ - 1}}\ln x - {x^{ - 1}}$ or equivalent

Use limits correctly, equate to $\frac{2}{5}$ and attempt rearrangement to obtain $\alpha $ in terms of $\ln \alpha $

Obtain given answer $\alpha  = \frac{5}{3}\left( {1 + \ln \alpha } \right)$ correctly

b) Use valid iterative formula correctly at least once

Obtain final answer $3.96$

Show sufficient iterations to > 4 dp to justify accuracy to 2 dp or show sign change in interval (3.955, 3.965)

$\left[ {4 \to 3.9772 \to 3.9676 \to 3.9636 \to 3.9619} \right]{\text{ }}$

SR: Use of ${\alpha _{n + 1}} = {e^{\left( {\frac{3}{5}{\alpha _n} - 1} \right)}}$ to obtain 0.50 also earns 3/3.

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!