Applications of Newton-Leibnitz Formula.

Here $f(t)=\log_e t$, $g(x)=x^2$, $h(x)=x^3$, so $g'=2x$ and $h'=3x^2$.

$\Phi'(x)=\log_e\!\left(x^{3}\right)\cdot 3x^{2}-\log_e\!\left(x^{2}\right)\cdot 2x = 9x^{2}\log_e x - 4x\log_e x = x\log_e x\,(9x-4).$

Check. Integrating first gives $\Phi(x)=\big[t\log_e t-t\big]_{x^2}^{x^3}=3x^3\log_e x-x^3-2x^2\log_e x+x^2$, and differentiating that returns $9x^2\log_e x+3x^2-3x^2-4x\log_e x-2x+2x = x\log_e x(9x-4)$. The two agree, as they must.

The integrand has no elementary antiderivative, so evaluating is not an option. Apply the formula with $h(x)=\sin x$:

$f'(x)=\sqrt{1+\sin^{4}x}\cdot\cos x.$

At $x=\dfrac{\pi}{3}$: $\sin x=\dfrac{\sqrt3}{2}$, so $\sin^4 x=\dfrac{9}{16}$, and $\cos x=\dfrac12$. Hence

$f'\!\left(\frac\pi3\right)=\sqrt{1+\frac{9}{16}}\cdot\frac12=\frac{5}{4}\cdot\frac12=\boxed{\dfrac58}.$

Step 1: Differentiate the given identity in $x$. The left side needs the chain rule:

$\frac{d}{dx}F\!\left(x^{2}\right)=F'\!\left(x^{2}\right)\cdot 2x.$

Step 2: Get $F'$ from Newton–Leibnitz. $F'(u)=u\,f(u)$, so $F'\!\left(x^{2}\right)=x^{2}f\!\left(x^{2}\right)$ and the left side is $2x^{3}f\!\left(x^{2}\right)$.

Step 3: Match.

$2x^{3}f\!\left(x^{2}\right)=4x^{3}+5x^{4}\ \Longrightarrow\ f\!\left(x^{2}\right)=2+\frac{5x}{2}.$

Step 4: Read the sum directly. Putting $x=r$ gives $f\!\left(r^{2}\right)=2+\dfrac{5r}{2}$, so

$\sum_{r=1}^{12}f\!\left(r^{2}\right)=\sum_{r=1}^{12}2+\frac52\sum_{r=1}^{12}r=24+\frac52\cdot\frac{12\cdot13}{2}=24+\frac52\cdot78=24+195.$

$\boxed{219}$

Note that $f$ itself is never written down as a function of its own argument — you only ever need $f$ at squares, and the substitution $x=r$ delivers exactly that.

Step 1: Spot the problem. The integrand contains $e^{x-t}$, which involves $x$. Newton–Leibnitz is illegal here.

Step 2: Split the exponential. $e^{x-t}=e^{x}\cdot e^{-t}$, and $e^x$ is constant with respect to $t$, so it comes out:

$f(x)=1-2x+e^{x}\int_0^x e^{-t}f(t)\,dt.$

Step 3: Isolate the integral. Multiply through by $e^{-x}$:

$e^{-x}f(x)=(1-2x)e^{-x}+\int_0^x e^{-t}f(t)\,dt.$

Now the integral is a clean function of the upper limit only.

Step 4: Differentiate. The left side needs the product rule; the integral gives $e^{-x}f(x)$:

$e^{-x}f'(x)-e^{-x}f(x)=-2e^{-x}-(1-2x)e^{-x}+e^{-x}f(x).$

Step 5: Divide by $e^{-x}$ and tidy.

$f'(x)-f(x)=-2-1+2x+f(x)\ \Longrightarrow\ f'(x)-2f(x)=2x-3.$

Step 6: Solve the linear ODE. Integrating factor $e^{-2x}$:

$\frac{d}{dx}\left(e^{-2x}f\right)=(2x-3)e^{-2x}.$

By parts, $\displaystyle\int(2x-3)e^{-2x}dx = -\frac{(2x-3)e^{-2x}}{2}-\frac{e^{-2x}}{2}=(1-x)e^{-2x}$, so $e^{-2x}f=(1-x)e^{-2x}+C$, i.e. $f(x)=1-x+Ce^{2x}$.

Step 7: Fix $C$. Put $x=0$ in the original relation: the integral vanishes, so $f(0)=1$. The formula gives $1+C$. Hence $C=0$ and

$\boxed{f(x)=1-x.}$

Step 1: Recognise the trap. $\sin(x-t)$ contains $x$. Differentiating blindly is wrong.

Step 2: Expand and pull the $x$-parts out. Treating $x$ as a constant for the $t$-integration,

$\sin(x-t)=\sin x\cos t-\cos x\sin t,$ $\int_0^x t^{2}\sin(x-t)\,dt=\sin x\int_0^x t^{2}\cos t\,dt-\cos x\int_0^x t^{2}\sin t\,dt.$

Step 3: Do the two standard by-parts integrals.

$\int_0^x t^{2}\cos t\,dt=\Big[t^{2}\sin t+2t\cos t-2\sin t\Big]_0^x = x^{2}\sin x+2x\cos x-2\sin x,$ $\int_0^x t^{2}\sin t\,dt=\Big[-t^{2}\cos t+2t\sin t+2\cos t\Big]_0^x = -x^{2}\cos x+2x\sin x+2\cos x-2.$

Step 4: Combine.

$\sin x\big(x^{2}\sin x+2x\cos x-2\sin x\big)-\cos x\big(-x^{2}\cos x+2x\sin x+2\cos x-2\big)$ $= x^{2}\big(\sin^{2}x+\cos^{2}x\big)+2x\sin x\cos x-2x\sin x\cos x-2\big(\sin^{2}x+\cos^{2}x\big)+2\cos x$ $= x^{2}-2+2\cos x.$

Step 5: Impose the condition. $x^{2}-2+2\cos x=x^{2}\Rightarrow\cos x=1\Rightarrow x=2k\pi$.

Step 6: Count in $[0,100]$. $2k\pi\le100\Rightarrow k\le\dfrac{100}{2\pi}\approx 15.9$, so $k=0,1,\dots,15$.

Answer: $\boxed{16}$

Step 1: The limits are constants, the $x$ is inside. Nothing to differentiate yet. Substitute.

Step 2: Let $u=\dfrac{tx}{36}$, so $dt=\dfrac{36}{x}du$; when $t=0$, $u=0$; when $t=36$, $u=x$. The $x$ has moved from the integrand into the upper limit:

$\int_0^{36}f\!\left(\frac{tx}{36}\right)dt=\frac{36}{x}\int_0^{x}f(u)\,du=4\alpha f(x).$

Step 3: Rearrange, then differentiate.

$\int_0^{x}f(u)\,du=\frac{\alpha x}{9}f(x)\ \xrightarrow{\ \frac{d}{dx}\ }\ f(x)=\frac{\alpha}{9}\big(f(x)+xf'(x)\big).$

Step 4: Separate. $(9-\alpha)f(x)=\alpha x f'(x)$, so $\dfrac{f'}{f}=\left(\dfrac{9}{\alpha}-1\right)\dfrac1x$ and $f(x)=c\,x^{\,9/\alpha-1}$.

Step 5: Use "standard parabola". $y=cx^{2}$ forces $\dfrac{9}{\alpha}-1=2$, so $\alpha=3$.

Step 6: Fit the point. $(2,1)$ gives $1=4c$, so $f(x)=\dfrac{x^{2}}{4}$ and $\beta=f(-4)=4$.

Answer: $\beta^{\alpha}=4^{3}=\boxed{64}$

Step 1: Confirm the form. As $x\to0$ the integral $\to0$ and $x^3\to0$: it is $\dfrac00$.

Step 2: L'Hôpital once. The derivative of the numerator is the integrand with $t$ replaced by $x$:

$\lim_{x\to0}\frac{\dfrac{x^{2}}{x^{4}+1}}{3x^{2}}=\lim_{x\to0}\frac{1}{3\left(x^{4}+1\right)}=\boxed{\dfrac13}.$

One application, no integration, no expansion.

$\displaystyle\int\sin(t^2)dt$ is a Fresnel integral — it has no elementary form, so this limit must be done by differentiation.

Step 1: L'Hôpital.

$\lim_{x\to0}\frac{\sin\!\left(x^{2}\right)}{3x^{2}}.$

Step 2: Standard limit. $\dfrac{\sin(x^{2})}{x^{2}}\to1$ as $x\to0$, so the value is $\boxed{\dfrac13}$.

Step 1: Numerator by Newton–Leibnitz with $h(x)=x^{2}$.

$\frac{d}{dx}\int_0^{x^{2}}\sec^{2}t\,dt=\sec^{2}\!\left(x^{2}\right)\cdot 2x.$

Step 2: Denominator. $\dfrac{d}{dx}(x\sin x)=\sin x+x\cos x$.

Step 3: Apply L'Hôpital.

$\lim_{x\to0}\frac{2x\sec^{2}\!\left(x^{2}\right)}{\sin x+x\cos x}.$

Still $\dfrac00$, but divide numerator and denominator by $x$ instead of differentiating again:

$\lim_{x\to0}\frac{2\sec^{2}\!\left(x^{2}\right)}{\dfrac{\sin x}{x}+\cos x}=\frac{2\cdot 1}{1+1}=\boxed{1}.$

This is precisely the definition of the derivative of $\Phi(x)=\displaystyle\int_1^x e^{t^2}dt$ at $x=1$, since $\Phi(1)=0$:

$\lim_{x\to1}\frac{\Phi(x)-\Phi(1)}{x-1}=\Phi'(1)=e^{1^{2}}=\boxed{e}.$

Recognising the difference quotient saves you even L'Hôpital.

Step 1: Confirm $\frac00$. At $x=\frac\pi2$ the limits of integration coincide, so the numerator vanishes; the denominator obviously does too.

Step 2: L'Hôpital, differentiating the numerator by Newton–Leibnitz. The upper limit is a constant, so it contributes nothing; the lower limit is $x^{3}$, and a moving lower limit carries a minus sign:

$\frac{d}{dx}\int_{x^{3}}^{(\pi/2)^{3}}(\cdots)dt=-\left(\sin\!\left(2x\right)+\cos x\right)\cdot 3x^{2},$

using $\left(x^{3}\right)^{1/3}=x$. The denominator differentiates to $2\left(x-\frac\pi2\right)$.

Step 3: Still $\frac00$. At $x=\frac\pi2$ we get $\sin\pi+\cos\frac\pi2=0+0=0$. Differentiate again:

$\frac{d}{dx}\Big[-3x^{2}\left(\sin2x+\cos x\right)\Big]=-6x\left(\sin2x+\cos x\right)-3x^{2}\left(2\cos2x-\sin x\right)$

over a denominator of $2$.

Step 4: Substitute $x=\frac\pi2$. The first term dies because $\sin\pi+\cos\frac\pi2=0$. For the second,

$2\cos\pi-\sin\frac\pi2=-2-1=-3,$

so the numerator is $-3\cdot\dfrac{\pi^{2}}{4}\cdot(-3)=\dfrac{9\pi^{2}}{4}$.

Step 5: Divide.

$\frac{9\pi^{2}/4}{2}=\boxed{\dfrac{9\pi^{2}}{8}}$

Numerical check. Evaluating the original quotient at $x=\frac\pi2\pm0.001$ gives $11.1127$ and $11.0939$, straddling $\frac{9\pi^{2}}{8}=11.1033$.

Step 1: Why two conditions, not one. For a finite limit after dividing by $x^{3}$, the bracket must vanish to order $x^{3}$. So the coefficient of $x$ must be zero and the coefficient of $x^{3}$ must equal $2$. That is the whole question.

Step 2: Expand the integral. By Newton–Leibnitz the integral $I(x)=\int_0^x\frac{dt}{1-t^2}$ has $I'(x)=\dfrac{1}{1-x^{2}}=1+x^{2}+x^{4}+\cdots$, and $I(0)=0$, so integrating term by term,

$I(x)=x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots$

Step 3: Expand the other term. $\beta x\cos x=\beta x\left(1-\dfrac{x^{2}}{2}+\cdots\right)=\beta x-\dfrac{\beta}{2}x^{3}+\cdots$

Step 4: Collect.

$\frac{\alpha}{2}I(x)+\beta x\cos x=\left(\frac{\alpha}{2}+\beta\right)x+\left(\frac{\alpha}{6}-\frac{\beta}{2}\right)x^{3}+O\!\left(x^{5}\right).$

Step 5: Impose both conditions.

$\frac{\alpha}{2}+\beta=0\ \Longrightarrow\ \alpha=-2\beta;\qquad \frac{\alpha}{6}-\frac{\beta}{2}=2.$

Substituting, $-\dfrac{\beta}{3}-\dfrac{\beta}{2}=-\dfrac{5\beta}{6}=2$, so $\beta=-\dfrac{12}{5}$ and $\alpha=\dfrac{24}{5}$.

Step 6:

$\alpha+\beta=\frac{24}{5}-\frac{12}{5}=\boxed{\dfrac{12}{5}=2.40}$

Step 1: Extract the constraint. The line $45x+5y+3=0$ is $y=-9x-\frac35$, so its slope is $-9$ and

$27r_1+\frac{9r_2}{2}=-9.$

Step 2: Check the form. The numerator vanishes at $x=3$ because the limits coincide. For the denominator $D(x)=\dfrac{3r_2}{2}x-r_2x^{2}-r_1x^{3}-3x$,

$D(3)=\frac{9r_2}{2}-9r_2-27r_1-9=-\left(27r_1+\frac{9r_2}{2}\right)-9=9-9=0.$

So the form is genuinely $\dfrac00$ — the constraint was placed there precisely to arrange it.

Step 3: L'Hôpital. The numerator differentiates by Newton–Leibnitz to $8x^{2}$, which at $x=3$ is $72$. For the denominator,

$D'(x)=\frac{3r_2}{2}-2r_2x-3r_1x^{2}-3,\qquad D'(3)=\frac{3r_2}{2}-6r_2-27r_1-3.$

Step 4: Use the constraint again. $\dfrac{3r_2}{2}-6r_2=-\dfrac{9r_2}{2}$, so

$D'(3)=-\left(27r_1+\frac{9r_2}{2}\right)-3=9-3=6.$

Step 5:

$\lim_{x\to3}\left(\int_3^x 8t^2\,dt\right)\Big/ D(x)=\frac{72}{6}=\boxed{12}$

Neither $r_1$ nor $r_2$ is ever determined individually — only the combination $27r_1+\frac{9r_2}{2}$ is needed, and it appears twice.

Step 1: Evaluate directly (here you can, and it is instructive). For $x\ge0$, $F(x)=\dfrac{x^{2}}{2}$; for $x<0$, $F(x)=-\dfrac{x^{2}}{2}$. So $F(x)=\dfrac{x|x|}{2}$.

Step 2: First derivative. By Newton–Leibnitz $F'(x)=|x|$, which exists for every $x$ including $0$ — the integrand $|t|$ is continuous, and continuity of the integrand is all the formula needs.

Step 3: Second derivative. $F''(x)=\dfrac{d}{dx}|x|$, which does not exist at $x=0$.

Conclusion: $F$ is differentiable on $\mathbb R$ but not twice differentiable at the origin. The corner of $|t|$ was not destroyed — it was pushed up one level.

Step 1: Continuity. $[\,t\,]$ is bounded on every bounded interval, so $|F(x+h)-F(x)|\le M|h|\to0$. Hence $F$ is continuous everywhere — in fact Lipschitz.

Step 2: Differentiability. Newton–Leibnitz gives $F'(x)=[\,x\,]$ wherever the integrand is continuous, i.e. for $x\notin\mathbb Z$. At an integer $n$ the one-sided derivatives are

$F'(n^{-})=n-1,\qquad F'(n^{+})=n,$

which differ by $1$. So $F$ is not differentiable at any integer.

Step 3: The value.

$F(2.5)=\int_0^{1}0\,dt+\int_1^{2}1\,dt+\int_2^{2.5}2\,dt = 0+1+1=\boxed{2}.$

Set $\Phi(u)=\displaystyle\int_2^{u}f(t)\,dt$, so $\Phi(2)=0$ and $\Phi'(u)=f(u)$. The limit is

$\lim_{h\to0}\frac{\Phi(2+3h)-\Phi(2)}{h}=3\lim_{h\to0}\frac{\Phi(2+3h)-\Phi(2)}{3h}=3\Phi'(2)=3f(2)=\boxed{15}.$

Step 1: Differentiate. $F'(x)=\dfrac{x\,e^{x}}{1+x^{2}}$.

Step 2: Read the sign. $e^{x}>0$ always and $1+x^{2}>0$ always, so the sign of $F'$ is exactly the sign of $x$.

Step 3: Conclude. $F'>0$ on $(0,\infty)$ and $F'<0$ on $(-\infty,0)$, with $F'(0)=0$. Hence $F$ decreases up to the origin and increases after it, so $x=0$ is the global minimum. $\blacksquare$

Step 1: $F'(x)=(x-1)(x-2)$.

Step 2: Sign chart. Both zeros are simple, so the sign alternates:

Step 3: Extrema. $+\to-$ at $x=1$ gives a local maximum; $-\to+$ at $x=2$ gives a local minimum.

Step 4: The values (this is the only place integration is needed). $F(x)=\dfrac{x^{3}}{3}-\dfrac{3x^{2}}{2}+2x$, so

$F(1)=\frac13-\frac32+2=\frac56,\qquad F(2)=\frac83-6+4=\frac23.$

Step 1: Build the gap function. Let $G(x)=x-\displaystyle\int_0^{x}\frac{dt}{\sqrt{1+t^{4}}}$, with $G(0)=0$.

Step 2: Differentiate. $G'(x)=1-\dfrac{1}{\sqrt{1+x^{4}}}$.

Step 3: Sign. For $x>0$ we have $1+x^{4}>1$, so $\sqrt{1+x^{4}}>1$, so $\dfrac{1}{\sqrt{1+x^{4}}}<1$ and $G'(x)>0$.

Step 4: Conclude. $G$ is strictly increasing on $[0,\infty)$ and $G(0)=0$, so $G(x)>0$ for $x>0$, which is the claim. $\blacksquare$

This "define the gap, differentiate, use monotonicity" pattern proves almost every integral inequality asked at JEE level.

Step 1: Differentiate. $f'(x)=(\sin x-\cos x)\left(e^{x}-2\right)(x-1)^{3}(x-2)^{5}$.

Step 2: All zeros in $(0,4]$.

• $\tan x=1\Rightarrow x=\dfrac\pi4\approx0.785$ and $x=\dfrac{5\pi}{4}\approx3.927$
• $e^{x}=2\Rightarrow x=\log_e2\approx0.693$
• $x=1$ (exponent 3) and $x=2$ (exponent 5)

Step 3: Check multiplicities. All five are odd, so $f'$ flips sign at every one of them.

Step 4: Anchor the sign. At $x=0.5$, left of all five: $\sin0.5-\cos0.5<0$, $e^{0.5}-2<0$, $(0.5-1)^3<0$, $(0.5-2)^5<0$. Four negatives multiply to a positive, so $f'(0.5)>0$.

Step 5: Alternate. In order the zeros are $0.693,\;0.785,\;1,\;2,\;3.927$, so the signs run

$+\;\big|\;-\;\big|\;+\;\big|\;-\;\big|\;+\;\big|\;-$

Step 6: Count $+\to-$. At $x=\log_e 2$, at $x=1$, and at $x=\dfrac{5\pi}{4}$.

Answer: $\boxed{3}$ local maxima.

Step 1: Use $f(t)=1-t$ from Example 4, so $f(t)+2=3-t$.

Step 2: Differentiate $g$.

$g'(x)=(3-x)^{15}(x-4)^{6}(x+12)^{17}=-(x-3)^{15}(x-4)^{6}(x+12)^{17},$

using $(3-x)^{15}=-(x-3)^{15}$ because $15$ is odd.

Step 3: Zeros and multiplicities. $x=-12$ (power 17, odd), $x=3$ (power 15, odd), $x=4$ (power 6, even).

Step 4: Sign chart.