please use this rule to answer the following question:
integrate u dv = uv - integrate v du
A function y= f(x) is defined by f(x) = x (0</= x </= e)
f(x) = ln (x) (e</= x </= e^2)
Find the area of the region bounded by the graph of f and the x-axis from x=0 to x= e^2.
the answer is 3/2 e^2, please answer this question steps by steps , thank you so much!!!
using integration
2009-05-19 6:21 am
回答 (2)
2009-05-19 7:03 am
2009-05-19 7:07 am
Area of the region bounded by the graph and x-axis
=∫f(x) dx (x from 0 to e)+∫f(x) dx (x from e to e^2)
=∫x dx (x from 0 to e)+∫lnx dx (x from e to e^2)
∫x dx (x from 0 to e)=[x^2/2] (x from 0 to e)=e^2/2-0=e^2/2
Let u= lnx
∫lnx dx (x from e to e^2)
=∫u dx (x from e to e^2)
=[ux] (u from 1 to 2) - ∫x du (u from 1 to 2)
=[xln(x)] (x from e to e^2) - ∫x 1/x dx (x from e to e^2)
{since du/dx= dln(x)/dx= 1/x}
=2e^2-e-∫dx (x from e to e^2)
=2e^2-e - [x] (x from e to e^2)
=2e^2- e -e^2 +e
=e^2
So the area of the bounded region
=e^2/2 +e^2
=3/2 e^2
=∫f(x) dx (x from 0 to e)+∫f(x) dx (x from e to e^2)
=∫x dx (x from 0 to e)+∫lnx dx (x from e to e^2)
∫x dx (x from 0 to e)=[x^2/2] (x from 0 to e)=e^2/2-0=e^2/2
Let u= lnx
∫lnx dx (x from e to e^2)
=∫u dx (x from e to e^2)
=[ux] (u from 1 to 2) - ∫x du (u from 1 to 2)
=[xln(x)] (x from e to e^2) - ∫x 1/x dx (x from e to e^2)
{since du/dx= dln(x)/dx= 1/x}
=2e^2-e-∫dx (x from e to e^2)
=2e^2-e - [x] (x from e to e^2)
=2e^2- e -e^2 +e
=e^2
So the area of the bounded region
=e^2/2 +e^2
=3/2 e^2
參考: ME
收錄日期: 2021-04-22 00:38:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090518000051KK01981