Mixed review R2: choosing a rate or accumulation method

A derivative value becomes useful after the point is found

The domain is all of $\R$, since $1+x^2>0$. The chain rule gives $f'=2x/(1+x^2)$. At $x=1$, the height is $\ln2$ and the slope is $1$, so the tangent line is $\boxed{y=\ln2+x-1}$. Substituting the input $1$ checks the point without confusing $\ln2$ with the intercept.

Match the inner derivative

The exponent is $x^2$ and its derivative $2x$ is available up to a constant factor. With $u=x^2$, $\dd u=2x\dd x$ and bounds $0,1$,

Integration by parts would require an antiderivative of $e^{x^2}$ for the naive choice $u=x$, or create a less convenient expression with another choice. The nested expression plus its derivative is the more useful signal here.

Make the logarithm simpler by differentiating it

Take $u=\ln x$, $\dd v=x\dd x$. Then $\dd u=\dd x/x$ and $v=x^2/2$. Parts gives

Differentiation produces $x\ln x+x/2-x/2=x\ln x$. The cancellation of the last two derivative terms checks the coefficient of the correction term.

Turn the infinite endpoint into a limit

For a finite $b>0$, the integral is $[\arctan x]_0^b=\arctan b$. As $b\to\infty$, this tends to $\boxed{\pi/2}$. Therefore the improper integral converges. The infinity symbol is an endpoint instruction to take a limit, not a number inserted into an antiderivative formula.

A shell’s radius and height have different roles

The parabola is nonnegative for $0\le x\le3$. A vertical strip rotates into a shell with radius $x$, height $3x-x^2$, and thickness $\dd x$. Hence

The integrand is nonnegative and has cubic-length units after integration. Replacing the radius by the height would describe a different geometry.

A rigorous error bound can be attained

Here $h=(1-(-1))/2=1$ and sample values are $1,0,1$. Thus $S_2=(1/3)(1+4\cdot0+1)=2/3$. The exact integral is $[x^5/5]_{-1}^1=2/5$, so the absolute error is $\boxed{4/15}$. Since $f^{(4)}=24$, the bound is

This quartic attains the bound. Exactness for cubics does not imply exactness for all smooth functions.

Shift to the equilibrium before separating

Set $z=y-5$. Then $z'=-2z$, so $z=Ce^{-2t}$. The initial value gives $C=-4$, hence $\boxed{y=5-4e^{-2t}}$. Its derivative is $8e^{-2t}$, while $-2(y-5)=8e^{-2t}$; also $y(0)=1$. As $t\to\infty$, $y$ increases toward $5$ from below and never reaches $5$ at a finite time. The constant equilibrium $y=5$ is another solution of the equation, but not of these initial data.

Continuity converts a pointwise fact into positive area

Assume instead that $f(c)>0$ somewhere in $[a,b]$. By continuity there is a nonzero-length subinterval next to or around $c$ on which $f(x)>f(c)/2$. Its integral is positive, and nonnegativity elsewhere prevents cancellation. Therefore the full integral would be positive, contradicting the premise. So $f(x)=0$ everywhere. Without continuity, a function equal to $1$ at one point and zero elsewhere is nonnegative and Riemann integrable with integral zero, yet is not identically zero. A single point contributes no positive-width area.