## Derivative of a Definite Integral

I have two equations as follows,

$$a = \int_0^1 dx \frac{c z_s^{d+1} x^d}{\sqrt{(1-(z_s/z_h)^{d+1} x^{d+1})(1-c^2 z_s^{2d} x^{2d})}} \tag{1}\label{1},$$

\begin{align} S &= \frac{1}{4 z_s^{d-1}}\Bigg(-\frac{\sqrt{(1-c^2 z_s^{2d})(1-b^{d+1})}}{d-1} – \frac{1}{d-1} c^2 z_s^{2d} \int^1_0 dx x^d \sqrt{\frac{(1-(b x)^{d+1})}{(1-c^2(z_s x)^{2d})}}\ & -\frac{b^{d+1}(d+1)}{2(d-1)} \int^1_0 dx x \sqrt{\frac{(1-c^2(z_s x)^{2d})}{(1-(b x)^{d+1})}}\ & + b^{d+1}\int^1_0 dx \frac{x}{\sqrt{(1-(b x)^{d+1})(1-c^2(z_s x)^{2d})}}\Bigg) \tag{2}\label{2} \end{align}

where $$c=c(z_s)$$ is a function of $$z_s$$, while $$a$$ (I can fix a value for this), $$z_h$$, $$d$$ (dimension) are constants, also $$b=z_s/z_h$$.

My goal is to obtain an expression for $$S$$ independent of $$c$$, the conditions that can help with this requirement are,

$$\frac{dS}{dz_s} = 0 \tag{3}\label{3},$$

and $$\eqref{1}$$. From $$\eqref{1}$$ and $$\eqref{2}$$, we can take the derivative with respect to $$z_s$$ and impose $$\eqref{3}$$ on the derivative of $$\eqref{2}$$ so that we can get expressions involving $$c$$ and $$\frac{dc}{dz_s}$$ in both the derivatives of $$\eqref{1}$$ and $$\eqref{2}$$, then eliminate both $$c$$ and $$\frac{dc}{dz_s}$$.

However, I tried to do the typical way as in the documentation but it does not produce the result I want.

d = 3; Integrate[(c[zs] zs^(d + 1) x^d)/((1 - x^(d + 1) (zs/zh)^(d + 1)) (1 - c[zs]^2 (zs x)^(2 d)))^(1/2), {x, 0, 1}] D[Integrate[(c[zs] zs^(d + 1) x^d)/((1 - x^(d + 1) (zs/zh)^(d + 1)) (1 - c[zs]^2 (zs x)^(2 d)))^(1/2), {x, 0, 1}] == 0, zs] 

## Error in nonlinearmodel fit for a function with Definite Integral and complex number

I am trying to fit a function and Here is the code which I am trying, Please find the data here dataset

Data = Import[    "E:\Shelender\codes\Mathematica\Aelastic \ relaxation\datat.asc"]; real = Data[[All, {1, 2}]]; imag = Data[[All, {1, 3}]]; w = 1.26*10^8; k = 1.38*10^-23;  f[H_, s_, d_] := ((1/(Sqrt[2*Pi]*s*H))*Exp[(-(Log[(H/d)])^2/(2*s^2))]) dynamic[x_?NumericQ, s_?NumericQ, d_?NumericQ, A_?NumericQ,    t_?NumericQ] :=   A*(1 - NIntegrate[      f[H, s, d]/((1 + (I*w*t*Exp[((H)/(k*x))]))), {H,        0, \[Infinity]}])  fit = ResourceFunction["MultiNonlinearModelFit"][    Rationalize[{real, imag}, 0],     ComplexExpand[ReIm@dynamic[x, s, d, A, t]],     Rationalize[{{A, 1.0*10^-4}, {t, 1.0*10^-12}, {d, 10}, {s, 0.25}},      0], {x}, PrecisionGoal -> 3, AccuracyGoal -> 3];  fit["ParameterTable"] Show[ListPlot[{real, imag}],   Plot[{fit[1, x], fit[2, x]}, {x, 0,     Max[real[[All, 1]], imag[[All, 1]]]}, PlotRange -> All],   PlotRange -> All] 

Although I am not getting any error but fit values are completely off

## How to calculate this kind of double definite integral directly

Let $$D=\left\{(x, y) \mid x^{2}+y^{2} \leq \sqrt{2}, x \geq 0, y \geq 0\right\}$$, $$\left[1+x^{2}+y^{2}\right]$$ represents the largest integer not greater than $$1+x^{2}+y^{2}$$, now I want to calculate this double integral $$\iint_{D} x y\left[1+x^{2}+y^{2}\right] d x d y$$.

reg = ImplicitRegion[x^2 + y^2 <= Sqrt[2] && x >= 0 && y >= 0, {x, y}]; Integrate[x*y*Round[1 + x^2 + y^2], {x, y} ∈ reg] 

But the result I calculated using the above method is not correct, the answer is $$\frac{3}{8}$$, what should I do to directly calculate this double integral (without using the technique of turning double integral into iterated integral)?

## How to use Trapezoidal Rule for approximating a definite integral from 0 to 30*60 5*v(t)\ dt

velocitydata = {200., 20.895780994409773, 10.986275727656292, 7.662410953506851, 5.998125878448737, 4.999000299900035, 4.332731606838267, 3.8567493313695387, 3.4997265945392857, 3.2220233380309256, 2.9998500112490625, 2.818065371605392, 2.666574078896326, 2.538386439381941, 2.428509477589222, 2.3332814832098125, 2.249956055974918, 2.176432934011989, 2.1110785329900605, 2.0526031497377395, 1.9999750004687404, 1.9523588169189425, 1.9090711873829898, 1.8695475468930842, 1.8333174191886965, 1.7999856001727976, 1.7692176833495532, 1.7407288016266662, 1.714274781445695, 1.6896451269910564, 1.6666574074845673, 1.645152730757279, 1.6249920654878016, 1.6060532320957523, 1.5882284246323966, 1.5714221574736715, 1.5555495542185434, 1.540534914053686, 1.526310504474162, 1.512815539733553, 1.4999953125219725, 1.4878004527119568, 1.4761862919948983, 1.4651123171705347, 1.4545416979860029, 1.4444408779281612, 1.4347792183895953, 1.4255286882592237, 1.4166635923132485, 1.408160332863668, 1.3999972000084, 1.3921541865573845, 1.3846128243129732, 1.3773560388842774, 1.3703680206329807, 1.3636341096975493, 1.3571406933360788, 1.350875114074985, 1.3448255873594115, 1.3389811275780437, 1.3333314814853396, 1.3278670681722977, 1.3225789248464297, 1.317458657775475, 1.3124983978300822, 1.3076907601301937, 1.3030288073598948, 1.2985060163674613, 1.2941162477124264, 1.2898537178607008, 1.285712973762941, 1.2816888695812, 1.2777765453550067, 1.2739714074208954, 1.27026911041953, 1.266665540742242, 1.2631568012844059, 1.259739197386856, 1.2564092238587217, 1.253163552985859, 1.2499990234386442, 1.2469126300014062, 1.2439015140533525, 1.240962954737622, 1.2380943607611097, 1.2352932627731243, 1.2325573062757642, 1.2298842450232272, 1.2272719348711656, 1.2247183280406884, 1.222221467764759, 1.2197794832875768, 1.2173905851900841, 1.2150530610170482, 1.2127652711832684, 1.2105256451383326, 1.2083326777710777, 1.2061849260364514, 1.204081005788894, 1.2020195888076308, 1.19999940000045, 1.1980192147735902, 1.1960778565563392, 1.1941741944698308, 1.192307141130332, 1.1904756505780585, 1.1886787163232235, 1.1869153695016552, 1.1851846771328776, 1.1834857404740797, 1.181817693463864, 1.180179701250117, 1.1785709587967457, 1.1769906895643907, 1.1754381442605732, 1.1739125996550557, 1.172413357456473, 1.170939743246574, 1.1694911054686528, 1.1680668144669837, 1.166666261574285, 1.1652888582444303, 1.1639340352278151, 1.162601241786949, 1.1612899449500012, 1.1599996288001782, 1.1587297937989383, 1.1574799561411815, 1.1562496471406645, 1.155038412644, 1.1538458124717037, 1.15267141988484, 1.1515148210759136, 1.150375614682726, 1.1492534113239998, 1.1481478331556445, 1.1470585134465943, 1.1459850961732263, 1.14492723563141, 1.1438845960653052, 1.142856851312065, 1.1418436844616617, 1.1408447875310836, 1.1398598611522026, 1.1388886142726473, 1.137930763869054, 1.1369860346721006, 1.1360541589027657, 1.1351348760192783, 1.1342279324742568, 1.1333330814815654, 1.1324500827924304, 1.1315787024803974, 1.1307187127347191, 1.1298698916617953, 1.1290320230942965, 1.1282048964076306, 1.1273883063434234, 1.1265820528397013, 1.1257859408674835, 1.1249997802735017, 1.1242233856287802, 1.1234565760828248, 1.1226991752231794, 1.1219510109401198, 1.1212119152962696, 1.120481724400925, 1.1197602782888993, 1.1190474208036916, 1.1183429994848042, 1.1176468654590366, 1.1169588733355942, 1.116278881104856, 1.1156067500406535, 1.114942344605917, 1.1142855323615606, 1.1136361838784714, 1.112994172652481, 1.1123593750222056, 1.1117316700896378, 1.1111109396433867};

## How definite of an order is there to a depth-first search of a graph?

I have the following graph that I need to simulate a depth-first search of; starting at g:

My question is: How definite of an order is there when performing a depth-first search? When doing a DFS of a tree, I always see the left-most child searched first (completely), then after backtracking, the second-most-left child…

But with a graph, “left” seems a lot more arbitrary.

For the above graph, I got the following order:

g, j, i, m, n, o, k, h, p, l, e, a, f, c, b, d `

But along the way, I found that there were many other possible paths to take. I’m guessing when implementing a DPS, I would visit the vertices in the order that they appear in the adjacency list (providing I’m using an adjacency list), but I don’t have such information here.

Am I right that there are many possible answers to this question? And is my trace of the DPS correct?

## how can I plot two variables (inside a definite integral) against each other

I want to plot x against y in this relation:

## How to find the original function from a definite integral and its value

What are the steps to solving this problem? (Thank you in advance)

The average value of a continuous function f (x) on the closed interval [3,7] is 18.

What is the value of (integral)[3,7] f (x)dx ?

## complex irrational definite integration

Finding value of $$\frac{8}{\pi}\int^{2}_{0}\frac{1}{x^2-x+1+\sqrt{x^4-4x^3+9x^2-10x+5}}$$

Let $$I=\int^{2}_{0}\frac{1}{x^2-x+1+\sqrt{x^4-4x^3+9x^2-10x+5}}dx$$

$$I=\int^{2}_{0}\frac{1}{x(x-1)+\sqrt{(x^2-2x)^2+5x(x-2)+5)}}dx$$

From $$\displaystyle \int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx$$

$$I=\int^{2}_{0}\frac{1}{(2-x)(1-x)+\sqrt{(2-x)^2x^2+5(2-x)(-x)+5}}dx$$

How do i solve it help me please

## Finding Limit involving definite integral

I am stuck in the following question..I tried to get Reduction formula and tried to use Sandwich Theorem but not reaching to final answer.

## Average approximates definite integral.

In my studies, I read a result similar to the following.

Let $$g:[0,1]\rightarrow \mathbb{R}$$ be twice continously differentiable on $$[0,1]$$ (this is probably stronger than necessary). Then $$$$\bigg\lvert 1/K_s \sum_{k=1}^{K_s} g\{x_i\} – \int_{0}^{1}g(u)du \bigg\rvert \leq \dfrac{C}{K_s},$$$$ for some constant $$C>0$$, where $$\{x_1,\dotsc,x_{K_s}\}$$ is a set of arbitrary equidistant points in $$[0,1]$$ (hence $$x_1=0$$ is not needed) and $$K_s$$ its cardinality.

I gave a cumbersome proof to this problem using the definition (upper/lower Riemann sum and constructing an appropriate partition). Though I think it is correct, I wonder if there is an easy way to prove it.

Question: what is the easiest way to prove it?