Can I have some help as to how this integrand be integrated? $$\int_{0}^{1}t^{n+1}\left ( u''v-v''u \right )dt+\int_{0}^{1}\left ( n+1 \right )t^{n}\left ( u'v-uv' \right )dt$$
$\begingroup$
$\endgroup$
2
-
$\begingroup$ notice that $(u''v - v''u) dt = d(u'v -uv')$ and then use integration by parts. $\endgroup$ – vnd Nov 3 '15 at 10:57
-
$\begingroup$ I get $$(u'v-uv')$$ as the end answer $\endgroup$ – integral Nov 3 '15 at 11:18
Add a comment
|
$\begingroup$
$\endgroup$
If we define $$f(t) = u'v - uv'$$ then $$f'(t) = u''v + u'v' - (u'v' + uv'')$$ $$ = u''v - uv''$$ This allows your integral to be expressed as $$I = \int_{0}^{1}t^{n+1}f'(t)dt + \int_{0}^{1}(n+1)t^nf(t)dt$$ Now we use integration by parts on the left hand integral to get $$\int_{0}^{1}t^{n+1}f'(t)dt = [t^{n+1}f(t)]_{0}^{1} - \int_{0}^{1}(n+1)t^nf(t)dt$$ and so substituting this back into the original expression for the integral we have $$I = [t^{n+1}f(t)]_{0}^{1} - \int_{0}^{1}(n+1)t^nf(t)dt + \int_{0}^{1}(n+1)t^nf(t)dt$$ which after cancellation is just $$I = [t^{n+1}f(t)]_{0}^{1}$$ $$ = f(1)$$ $$ = u'(1)v(1) - u(1)v'(1)$$
