Question Number 1602 by 112358 last updated on 25/Aug/15
$${Show}\:{that}\: \\ $$$$\frac{{d}}{{dy}}\int_{{g}_{\mathrm{1}} \left({y}\right)} ^{{g}_{\mathrm{2}} \left({y}\right)} {f}\left({x},{y}\right){dx}=\int_{{g}_{\mathrm{1}} \left({y}\right)} ^{{g}_{\mathrm{2}} \left({y}\right)} \frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dx}+{g}_{\mathrm{2}} ^{'} \left({y}\right){f}\left({g}_{\mathrm{2}} \left({y}\right),{y}\right)−{g}_{\mathrm{1}} ^{'} \left({y}\right){f}\left({g}_{\mathrm{1}} \left({y}\right),{y}\right) \\ $$$${using}\:{Leibniz}'{s}\:{rule}\:{and}\:{the} \\ $$$${chain}\:{rule}\:{where}\:{g}_{\mathrm{1}} \:{and}\:{g}_{\mathrm{2}} \:{are} \\ $$$${differentiable}. \\ $$
Commented by 123456 last updated on 26/Aug/15
![(d/dy)∫_(g_1 (y)) ^(g_2 (y)) f(x)dx=f[g_2 (y)]g′_2 (y)−f[g_1 (y)]g′_1 (y) ∫_(g_1 (y)) ^(g_2 (y)) f(x)dx=F[g_2 (y)]−F[g_1 (y)]](https://www.tinkutara.com/question/Q1606.png)
$$\frac{{d}}{{dy}}\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x}\right){dx}={f}\left[{g}_{\mathrm{2}} \left({y}\right)\right]{g}'_{\mathrm{2}} \left({y}\right)−{f}\left[{g}_{\mathrm{1}} \left({y}\right)\right]{g}'_{\mathrm{1}} \left({y}\right) \\ $$$$\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x}\right){dx}=\mathrm{F}\left[{g}_{\mathrm{2}} \left({y}\right)\right]−\mathrm{F}\left[{g}_{\mathrm{1}} \left({y}\right)\right] \\ $$
Commented by 112358 last updated on 26/Aug/15
![I got what you′ve commented but I don′t know how the term ∫_(g_1 (y)) ^(g_2 (y) (∂f/∂y)(x,y)dx appears. Leibniz′s rule gives (d/dy)∫_a ^b f(x,y)dx=∫_a ^b (∂f/∂y)(x,y)dx for f being a function on the rectangle R=[a,b]×[c,d] with (∂f/∂y)(x,y) being continuous on R.](https://www.tinkutara.com/question/Q1607.png)
$${I}\:{got}\:{what}\:{you}'{ve}\:{commented}\:{but}\:{I} \\ $$$${don}'{t}\:{know}\:{how}\:{the}\:{term}\:\int_{{g}_{\mathrm{1}} \left({y}\right)} ^{{g}_{\mathrm{2}} \left({y}\right.} \frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dx} \\ $$$${appears}. \\ $$$${Leibniz}'{s}\:{rule}\:{gives} \\ $$$$\frac{{d}}{{dy}}\int_{{a}} ^{{b}} {f}\left({x},{y}\right){dx}=\int_{{a}} ^{{b}} \frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dx} \\ $$$${for}\:{f}\:{being}\:{a}\:{function}\:{on} \\ $$$${the}\:{rectangle}\:{R}=\left[{a},{b}\right]×\left[{c},{d}\right]\:{with} \\ $$$$\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:{being}\:{continuous}\:{on}\:{R}. \\ $$
Commented by 123456 last updated on 28/Aug/15
$$\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dx}=\underset{\mathrm{0}} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dy}−\underset{\mathrm{0}} {\overset{{g}_{\mathrm{1}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dx} \\ $$$${h}\left({y}\right)=\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dx} \\ $$
Answered by 123456 last updated on 29/Aug/15
![h(y)=∫_(g_1 (y)) ^(g_2 (y)) f(x,y)dx (dh/dy)=lim_(Δy→0) ((h(y+Δy)−h(y))/(Δy)) =lim_(Δy→0) (1/(Δy))[∫_(g_1 (y+Δy)) ^(g_2 (y+Δy)) f(x,y+Δy)dx−∫_(g_1 (y)) ^(g_2 (y)) f(x,y)dx] =lim_(Δy→0) (1/(Δy))[∫_(g_2 (y)) ^(g_2 (y+Δy)) f(x,y+Δy)dx+∫_(g_1 (y)) ^(g_2 (y)) f(x,y+Δy)dx+∫_(g_1 (y+Δy)) ^(g_1 (y)) f(x,y+Δy)dx−∫_(g_1 (y)) ^(g_2 (y)) f(x,y)dx] =lim_(Δy→0) (1/(Δy))∫_(g_2 (y)) ^(g_2 (y+Δy)) f(x,y+Δy)dx+lim_(Δy→0) (1/(Δy))∫_(g_1 (y+Δy)) ^(g_1 (y)) f(x,y+Δy)dx+lim_(Δy→0) (1/(Δy))∫_(g_1 (y)) ^(g_2 (y)) f(x,y+Δy)−f(x,y)dx =lim_(Δy→0) (1/(Δy))∫_(g_2 (y)) ^(g_2 (y+Δy)) f(x,y+Δy)dx+lim_(Δy→0) (1/(Δy))∫_(g_1 (y+Δy)) ^(g_1 (y)) f(x,y+Δy)dx+∫_(g_1 (y)) ^(g_2 (y)) lim_(Δy→0) ((f(x,y+Δy)−f(x,y))/(Δy))dx =lim_(Δy→0) (1/(Δy))∫_(g_2 (y)) ^(g_2 (y+Δy)) f(x,y+Δy)dx+lim_(Δy→0) (1/(Δy))∫_(g_1 (y+Δy)) ^(g_1 (y)) f(x,y+Δy)dx+∫_(g_1 (y)) ^(g_2 (y)) (∂f/∂y)(x,y)dx by mean value theorem ∃ξ_2 ∈[g_2 (y),g_2 (y+Δy)]⇒∫_(g_2 (y)) ^(g_2 (y+Δy)) f(x,y+Δy)dx=[g_2 (y+Δy)−g_2 (y)]f(ξ_1 ,y+Δy) ∃ξ_1 ∈[g_1 (y+Δy),g_1 (y)]⇒∫_(g_1 (y+Δy)) ^(g_1 (y)) f(x,y+Δy)dx=[g_1 (y)−g_1 (y+Δy)]f(ξ_1 ,y+Δy) (dh/dy)=lim_(Δy→0) ((g_2 (y+Δy)−g_2 (y))/(Δy))f(ξ_2 ,y+Δy)−lim_(Δy→0) ((g_1 (y+Δy)−g_1 (y))/(Δy))f(ξ_1 ,y+Δy)+∫_(g_1 (y)) ^(g_2 (y)) (∂f/∂y)(x,y)dx as Δy→0,ξ_1 →g_1 (y),ξ_2 →g_2 (y) (dh/dy)=lim_(Δy→0) ((g_2 (y+Δy)−g_2 (y))/(Δy))lim_(Δy→0) f[g_2 (y),y+Δy]−lim_(Δy→0) ((g_1 (y+Δy)−g_1 (y))/(Δy))lim_(Δy→0) f[g_1 (y),y+Δy]+∫_(g_1 (y)) ^(g_2 (y)) (∂f/∂y)(x,y)dxy (dh/dy)=g′_2 (y)f[g_2 (y),y]−g′_1 (y)f[g_1 (y),y]+∫_(g_1 (y)) ^(g_2 (y)) (∂f/∂y)(x,y)dx](https://www.tinkutara.com/question/Q1646.png)
$${h}\left({y}\right)=\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dx} \\ $$$$\frac{{dh}}{{dy}}=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{h}\left({y}+\Delta{y}\right)−{h}\left({y}\right)}{\Delta{y}} \\ $$$$\:\:\:\:\:\:=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\left[\underset{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}−\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dx}\right] \\ $$$$\:\:\:\:\:\:=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\left[\underset{{g}_{\mathrm{2}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)} {\overset{{g}_{\mathrm{1}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}−\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}\right){dx}\right] \\ $$$$\:\:\:\:\:\:=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{2}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)} {\overset{{g}_{\mathrm{1}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right)−{f}\left({x},{y}\right){dx} \\ $$$$\:\:\:\:\:\:=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{2}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)} {\overset{{g}_{\mathrm{1}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x},{y}+\Delta{y}\right)−{f}\left({x},{y}\right)}{\Delta{y}}{dx} \\ $$$$\:\:\:\:\:\:=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{2}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\Delta{y}}\underset{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)} {\overset{{g}_{\mathrm{1}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}+\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}\frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dx} \\ $$$$\mathrm{by}\:\mathrm{mean}\:\mathrm{value}\:\mathrm{theorem} \\ $$$$\exists\xi_{\mathrm{2}} \in\left[{g}_{\mathrm{2}} \left({y}\right),{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)\right]\Rightarrow\underset{{g}_{\mathrm{2}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}=\left[{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)−{g}_{\mathrm{2}} \left({y}\right)\right]{f}\left(\xi_{\mathrm{1}} ,{y}+\Delta{y}\right) \\ $$$$\exists\xi_{\mathrm{1}} \in\left[{g}_{\mathrm{1}} \left({y}+\Delta{y}\right),{g}_{\mathrm{1}} \left({y}\right)\right]\Rightarrow\underset{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)} {\overset{{g}_{\mathrm{1}} \left({y}\right)} {\int}}{f}\left({x},{y}+\Delta{y}\right){dx}=\left[{g}_{\mathrm{1}} \left({y}\right)−{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)\right]{f}\left(\xi_{\mathrm{1}} ,{y}+\Delta{y}\right) \\ $$$$\frac{{dh}}{{dy}}=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)−{g}_{\mathrm{2}} \left({y}\right)}{\Delta{y}}{f}\left(\xi_{\mathrm{2}} ,{y}+\Delta{y}\right)−\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)−{g}_{\mathrm{1}} \left({y}\right)}{\Delta{y}}{f}\left(\xi_{\mathrm{1}} ,{y}+\Delta{y}\right)+\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}\frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dx} \\ $$$$\mathrm{as}\:\Delta{y}\rightarrow\mathrm{0},\xi_{\mathrm{1}} \rightarrow{g}_{\mathrm{1}} \left({y}\right),\xi_{\mathrm{2}} \rightarrow{g}_{\mathrm{2}} \left({y}\right) \\ $$$$\frac{{dh}}{{dy}}=\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{g}_{\mathrm{2}} \left({y}+\Delta{y}\right)−{g}_{\mathrm{2}} \left({y}\right)}{\Delta{y}}\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left[{g}_{\mathrm{2}} \left({y}\right),{y}+\Delta{y}\right]−\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{g}_{\mathrm{1}} \left({y}+\Delta{y}\right)−{g}_{\mathrm{1}} \left({y}\right)}{\Delta{y}}\underset{\Delta{y}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left[{g}_{\mathrm{1}} \left({y}\right),{y}+\Delta{y}\right]+\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}\frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dxy} \\ $$$$\frac{{dh}}{{dy}}={g}'_{\mathrm{2}} \left({y}\right){f}\left[{g}_{\mathrm{2}} \left({y}\right),{y}\right]−{g}'_{\mathrm{1}} \left({y}\right){f}\left[{g}_{\mathrm{1}} \left({y}\right),{y}\right]+\underset{{g}_{\mathrm{1}} \left({y}\right)} {\overset{{g}_{\mathrm{2}} \left({y}\right)} {\int}}\frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dx} \\ $$
Commented by 123456 last updated on 29/Aug/15
![mean value theorem for integrais if f is integrable and continuous on [a,b] ∃ξ∈[a,b] ∫_a ^b f(x)dx=f(ξ)(b−a)](https://www.tinkutara.com/question/Q1649.png)
$$\mathrm{mean}\:\mathrm{value}\:\mathrm{theorem}\:\mathrm{for}\:\mathrm{integrais} \\ $$$$\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{integrable}\:\mathrm{and}\:\mathrm{continuous}\:\mathrm{on}\:\left[{a},{b}\right] \\ $$$$\exists\xi\in\left[{a},{b}\right] \\ $$$$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}={f}\left(\xi\right)\left({b}−{a}\right) \\ $$