Leibnitz's rule, Wall's Formula Questions in English

(A) Let $L = \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$.
Using the property $\int_{b}^{a} f(t) dt = -\int_{a}^{b} f(t) dt$,we can rewrite the expression as:
$L = \lim _{x}$ ${\rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t + \int_{a}^{x+y} e^{\sin ^{2} t} d t\right]$
$L = \lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$.
Since this is a $\frac{0}{0}$ form,we apply $L$'$H$ôpital's Rule and the Leibniz Integral Rule:
$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx} \int_{y}^{x+y} e^{\sin ^{2} t} d t}{\frac{d}{dx} (x)}$
$L = \lim _{x}$ ${\rightarrow 0} \frac{e^{\sin ^{2}(x+y)} \cdot \frac{d}{dx}(x+y) - e^{\sin ^{2}(y)} \cdot \frac{d}{dx}(y)}{1}$
$L = \lim _{x \rightarrow 0} e^{\sin ^{2}(x+y)} \cdot (1) - 0 = e^{\sin ^{2} y}$.

(A) Let $L = \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L$' Hospital's rule.
Using the Leibniz rule for differentiation under the integral sign,the derivative of the numerator is $\cos(x^4) \cdot \frac{d}{dx}(x^2) = 2x \cos(x^4)$.
The derivative of the denominator $x \sin x$ is $\sin x + x \cos x$.
So,$L = \lim _{x \rightarrow 0} \frac{2x \cos(x^4)}{\sin x + x \cos x}$.
Divide numerator and denominator by $x$:
$L = \lim _{x \rightarrow 0} \frac{2 \cos(x^4)}{\frac{\sin x}{x} + \cos x}$.
As $x \rightarrow 0$,$\frac{\sin x}{x} \rightarrow 1$,$\cos(x^4) \rightarrow 1$,and $\cos x \rightarrow 1$.
Therefore,$L = \frac{2(1)}{1 + 1} = \frac{2}{2} = 1$.

(C) Given $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$.
Using the Leibniz integral rule,we differentiate $g(x)$ with respect to $x$:
$g'(x) = \frac{f(2x)}{2x} \cdot \frac{d}{dx}(2x) - \frac{f(x)}{x} \cdot \frac{d}{dx}(x)$
$g'(x) = \frac{f(2x)}{2x} \cdot 2 - \frac{f(x)}{x} \cdot 1$
$g'(x) = \frac{f(2x)}{x} - \frac{f(x)}{x}$
Since it is given that $f(2x) = f(x)$,we substitute this into the expression:
$g'(x) = \frac{f(x) - f(x)}{x} = 0$
Since $g'(x) = 0$ for all $x > 0$,the function $g(x)$ is a constant function.

(C) To evaluate the integral $I = \int_0^{\pi / 2} \sin^5 x \, dx$,we use the Wallis formula for $\int_0^{\pi / 2} \sin^n x \, dx$,which is $\frac{(n-1)!!}{n!!} \times \frac{\pi}{2}$ if $n$ is even,and $\frac{(n-1)!!}{n!!}$ if $n$ is odd.
For $n = 5$ (which is odd),the formula gives:
$I = \frac{(5-1) \times (5-3)}{5 \times 3 \times 1} = \frac{4 \times 2}{5 \times 3 \times 1} = \frac{8}{15}$.
Alternatively,using substitution:
$I = \int_0^{\pi / 2} \sin^4 x \sin x \, dx = \int_0^{\pi / 2} (1 - \cos^2 x)^2 \sin x \, dx$.
Let $u = \cos x$,then $du = -\sin x \, dx$. When $x = 0, u = 1$; when $x = \pi / 2, u = 0$.
$I = -\int_1^0 (1 - u^2)^2 \, du = \int_0^1 (1 - 2u^2 + u^4) \, du$.
$I = [u - \frac{2u^3}{3} + \frac{u^5}{5}]_0^1 = 1 - \frac{2}{3} + \frac{1}{5} = \frac{15 - 10 + 3}{15} = \frac{8}{15}$.

(A) Let $f(x) = \int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$.
To find the points of extremum,we calculate the derivative $f'(x)$ using the Leibniz Integral Rule:
$f'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot \frac{d}{dx}(x^2)$.
$f'(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot (2x)$.
$f'(x) = \frac{(x^2-1)(x^2-4)}{2 + e^{x^2}} \cdot (2x)$.
$f'(x) = \frac{(x-1)(x+1)(x-2)(x+2)(2x)}{2 + e^{x^2}}$.
Setting $f'(x) = 0$,we get $x = 0, 1, -1, 2, -2$.
Thus,the points of extremum are $0, \pm 1, \pm 2$.

(B) Using the Leibniz integral rule,the derivative of $f(x) = \int_{g(x)}^{h(x)} F(t) dt$ is given by $f^{\prime}(x) = F(h(x)) \cdot h^{\prime}(x) - F(g(x)) \cdot g^{\prime}(x)$.
Here,$F(t) = e^{-t^2}$,$h(x) = \cos x$,and $g(x) = \sin x$.
So,$f^{\prime}(x) = e^{-(\cos x)^2} \cdot (-\sin x) - e^{-(\sin x)^2} \cdot (\cos x)$.
At $x = \frac{\pi}{4}$,we have $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Substituting these values:
$f^{\prime}\left(\frac{\pi}{4}\right) = -e^{-1/2} \cdot \frac{1}{\sqrt{2}} - e^{-1/2} \cdot \frac{1}{\sqrt{2}}$
$f^{\prime}\left(\frac{\pi}{4}\right) = -\frac{2}{\sqrt{2} \cdot \sqrt{e}} = -\frac{\sqrt{2}}{\sqrt{e}} = -\sqrt{\frac{2}{e}}$.

(C) Given the equation: $\int_0^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$.
Let $u = \frac{tx}{36}$,then $du = \frac{x}{36} dt$,so $dt = \frac{36}{x} du$.
When $t=0, u=0$ and when $t=36, u=x$.
The integral becomes: $\int_0^x f(u) \cdot \frac{36}{x} du = 4\alpha f(x)$.
$\int_0^x f(u) du = \frac{4\alpha x f(x)}{36} = \frac{\alpha x f(x)}{9}$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$f(x) = \frac{\alpha}{9} [f(x) + x f'(x)]$.
$f(x) = \frac{\alpha}{9} f(x) + \frac{\alpha x}{9} f'(x)$.
$(1 - \frac{\alpha}{9}) f(x) = \frac{\alpha x}{9} f'(x) \Rightarrow (9 - \alpha) f(x) = \alpha x f'(x)$.
$\frac{f'(x)}{f(x)} = \frac{9 - \alpha}{\alpha} \cdot \frac{1}{x}$.
Integrating both sides: $\ln|f(x)| = (\frac{9}{\alpha} - 1) \ln|x| + C$.
$f(x) = c x^{(\frac{9}{\alpha} - 1)}$.
Since $f(x)$ is a standard parabola,the exponent must be $2$,so $\frac{9}{\alpha} - 1 = 2 \Rightarrow \frac{9}{\alpha} = 3 \Rightarrow \alpha = 3$.
Thus,$f(x) = cx^2$.
Since it passes through $(2, 1)$,$1 = c(2)^2 \Rightarrow c = \frac{1}{4}$.
So,$f(x) = \frac{x^2}{4}$.
For the point $(-4, \beta)$,$\beta = \frac{(-4)^2}{4} = \frac{16}{4} = 4$.
Finally,$\beta^{\alpha} = 4^3 = 64$.