# Leibniz Integ R Derivative Integration, Differentiation And Integration In R

**Leibniz Integ R Derivative Integration, Differentiation And Integration In R**in

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$$-frac{1}{2}left(frac{3+sqrt{r})^2}{2} + (3 + sqrt{r})sqrt{r}+3(3+sqrt{r})

ight)-frac{1}{2}left(-frac{(2^2)}{2}+2sqrt{r}+3cdot 2

ight)$$

After I simplified I got:

$$frac{1 + 2sqrt{r}+r}{4}$$

I should get:

$$frac{1 + sqrt{r}}{4sqrt{r}}$$

But I cannot get this result.Can someone help? what am I missing?

Đang xem: R derivative integration

As it appears the derivative is taken with respect to $r$. The integral in question is given by, and evaluated as, the following:egin{align}I(r) &= frac{1}{2} , int_{2}^{3+sqrt{r}} (3 + sqrt{r} – t) , dt \&= frac{1}{2} , left< (3 + sqrt{r}) t - frac{t^{2}}{2} ight>_{2}^{3 + sqrt{r}} \I(r) &= frac{(3 + sqrt{r})^{2}}{4} – (2 + sqrt{r}).end{align}Now by differentiation egin{align}frac{dI}{dr} &= frac{2}{4} cdot (3 + sqrt{r}) cdot frac{1}{2 sqrt{r}} – frac{1}{2 sqrt{r}} \&= frac{3 + sqrt{r}}{4 sqrt{r}} – frac{1}{2 sqrt{r}} = frac{1 + sqrt{r}}{4 , sqrt{r}}.end{align}

I think that the obvious purpose of that problem is to use differentiation under the integral sign. As seen on wikipedia, the theorem states

$$frac{wnyrails.orgrm{d}}{wnyrails.orgrm{d}x} left (int_{a(x)}^{b(x)}f(x,t),wnyrails.orgrm{d}t

ight) = f(x,b(x))cdot b”(x) – f(x,a(x))cdot a”(x) + int_{a(x)}^{b(x)} frac{partial}{partial x}f(x,t); wnyrails.orgrm{d}t$$(So in your notation $x=r$ and $t=c$). In this case this becomes particularly simple, because $f(x, b(x)) = 3+sqrt{x}- (3+sqrt{x})=0$ and $a”(x)=0$ hence we are left with $$int_{a(x)}^{b(x)} frac{partial}{partial x}f(x,t); wnyrails.orgrm{d}t = frac{1}{4sqrt x}int_{2}^{3+sqrt{x}} 1; wnyrails.orgrm{d}t = frac{1+sqrt{x}}{4sqrt{x}}$$

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$$int_{2}^{3+sqrt{r}} left(frac{1}{2}left(3+sqrt{r}-c

ight)

ight)dc=$$$$frac{1}{2}cdotint_{2}^{3+sqrt{r}} left(3+sqrt{r}-c

ight)dc=$$$$frac{1}{2}left(int_{2}^{3+sqrt{r}} (3)dc+int_{2}^{3+sqrt{r}} (sqrt{r}) dc-int_{2}^{3+sqrt{r}} (c)dc

ight)=$$$$frac{1}{2}left(left<3c
ight>_{2}^{3+sqrt{r}}+left

ight)=$$

$$frac{1}{2}left(left(3sqrt{r}+3

ight)+left(r+sqrt{r}

ight)-left(frac{r}{2}+3sqrt{r}+frac{5}{2}

ight)

ight)=$$

$$frac{1}{2}left(frac{1}{2}left(sqrt{r}+1

ight)^2

ight)=$$

$$frac{1}{4}left(sqrt{r}+1

ight)^2=frac{left(sqrt{r}+1

ight)^2}{4}$$

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