Given ddx-ux-vx then -ab uxvxdx

Given, #160; ddx[u(x)]=v(x) or, d[u(x)]=v(x)dx ......(1).Now, ∫_a^b u(x)v(x)dx =∫_a^b u(x)d[u(x)] [ Using (1)] =[[u(x)]^22]_x=a^x=b ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Given ddx[u...

Question

Given $\frac{d}{d x} [u (x)] = v (x)$ then $b \int a u (x) v (x) d x$

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Solution

Given,
$\frac{d}{d x} [u (x)] = v (x)$
or, $d [u (x)] = v (x) d x$ ......(1).
Now,
$b \int a u (x) v (x) d x$
$= b \int a u (x) d [u (x)]$ [ Using (1)]
$= {[\frac{[u (x)]^{2}}{2}]}_{x = a}^{x = b}$
$= \frac{1}{2} ([u (a)]^{2} - [u (b)]^{2})$

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Given ddx[u(x)]=v(x) then b∫au(x)v(x)dx

Given, ddx[u(x)]=v(x)or, d[u(x)]=v(x)dx......(1).Now,b∫au(x)v(x)dx=b∫au(x)d[u(x)] [ Using (1)]=[[u(x)]22]x=bx=a=12([u(a)]2−[u(b)]2)

Given $\frac{d}{d x} [u (x)] = v (x)$ then $b \int a u (x) v (x) d x$

Given,
$\frac{d}{d x} [u (x)] = v (x)$
or, $d [u (x)] = v (x) d x$ ......(1).
Now,
$b \int a u (x) v (x) d x$
$= b \int a u (x) d [u (x)]$ [ Using (1)]
$= {[\frac{[u (x)]^{2}}{2}]}_{x = a}^{x = b}$
$= \frac{1}{2} ([u (a)]^{2} - [u (b)]^{2})$