Self-check: 201-SN3-RE, Integral Calculus

a) General method: divide top and bottom by the highest power present, here $x^{2}$: $f(x)=\frac{3-\frac{2}{x}+\frac{1}{x^{2}}}{1-\frac{4}{x^{2}}}$. As $x\to\pm\infty$, all the terms in $\frac{1}{x}$ and $\frac{1}{x^{2}}$ vanish: the limit is $\frac{3}{1}=3$ on BOTH sides. Horizontal asymptote: $y=3$, the horizontal dashed line in the solution figure. A legitimate shortcut for a rational function with EQUAL degrees: the ratio of the leading coefficients, here $\frac{3}{1}$.

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a) Three layers: the exponential, then the sine, then $2x$. $f'(x)=e^{\sin(2x)}\cdot\cos(2x)\cdot 2=2\cos(2x)e^{\sin(2x)}$. The factor 2 from the innermost layer is the one that gets forgotten: check at $x=0$, $f'(0)=2\cdot 1\cdot e^{0}=2$, whereas without the factor you would find 1, a slope half as large as it should be.

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a) For $x\geq 0$, $|x|=x$, so an antiderivative is $\frac{x^{2}}{2}+C_{1}$. For $x<0$, $|x|=-x$, so an antiderivative is $-\frac{x^{2}}{2}+C_{2}$. These two pieces must JOIN UP: an antiderivative is differentiable, hence continuous, and continuity at 0 forces $\frac{0}{2}+C_{1}=-\frac{0}{2}+C_{2}$, that is, $C_{1}=C_{2}$. So there is only one constant, not two: $F(x)=\frac{x^{2}}{2}+C$ for $x\geq 0$, $F(x)=-\frac{x^{2}}{2}+C$ for $x<0$.

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a) Exclude the roots of the denominator. $2x-5=0$ gives $x=\frac52$, so the domain is $\mathbb{R}\setminus\left\{\frac52\right\}$. $x^2+9$ never vanishes on $\mathbb{R}$, so the domain is all of $\mathbb{R}$: an irreducible denominator creates no excluded value, which is the first visible sign that it is irreducible. $2x^2-8x=2x(x-4)$ vanishes at 0 and at 4, so the domain is $\mathbb{R}\setminus\{0, 4\}$. Finally $(x-3)^2$ vanishes only at 3, domain $\mathbb{R}\setminus\{3\}$.

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a) $x^{2}+x-2=(x+2)(x-1)$. Set $\frac{x+5}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}$, that is $x+5=A(x-1)+B(x+2)$. Method of convenient values: $x=1$: $6=3B$, $B=2$; $x=-2$: $3=-3A$, $A=-1$. Hence $\int=-\ln|x+2|+2\ln|x-1|+C=\ln\frac{(x-1)^{2}}{|x+2|}+C$.

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d) The substitution requires $t=\tan\frac{x}{2}$ to be continuous, so $\frac{x}{2}$ must avoid $\frac{\pi}{2}$: it is only legitimate on an interval contained in $\left(-\pi,\pi\right)$, up to a shift by a multiple of $2\pi$. On $[0,2\pi]$, $t$ starts at 0, tends to $+\infty$ at $x=\pi$, reappears at $-\infty$ and climbs back to 0: blindly applying the antiderivative $\dfrac{2}{\sqrt{3}}\arctan\dfrac{t}{\sqrt{3}}$ gives $F\left(2\pi\right)-F\left(0\right)=0-0=0$. Yet the integrand $\dfrac{1}{2+\cos x}$ is everywhere greater than $\frac{1}{3}$: the integral is at least $\frac{2\pi}{3}$. The true value, obtained by splitting at $x=\pi$ and doubling by symmetry, is $\dfrac{2\pi}{\sqrt{3}}\approx 3.628$. Remember: a zero for the integral of a strictly positive function is not a computation error, it is a DOMAIN error.

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a) As $x\to+\infty$: the oblique asymptote is written $y=mx+p$ with $m=\lim\frac{f(x)}{x}$ and $p=\lim\left(f(x)-mx\right)$. Here $\frac{\sqrt{x^{2}+3x}}{x}=\sqrt{1+\frac{3}{x}}\to 1$, so $m=1$; then $p=\lim_{x\to\infty}\left(\sqrt{x^{2}+3x}-x\right)=\frac{3}{2}$ by the conjugate, a computation already done in exercise 3d. The asymptote is $y=x+\frac{3}{2}$. As $x\to-\infty$: careful, $\sqrt{x^{2}}=|x|=-x$ since $x$ is negative. Then $\frac{\sqrt{x^{2}+3x}}{x}=\frac{|x|\sqrt{1+\frac{3}{x}}}{x}=-\sqrt{1+\frac{3}{x}}\to -1$, so $m=-1$; and $p=\lim_{x\to-\infty}\left(\sqrt{x^{2}+3x}+x\right)=-\frac{3}{2}$ by the same conjugate. The asymptote is $y=-x-\frac{3}{2}$. The difference comes ENTIRELY from the simplification $\sqrt{x^{2}}=|x|$: writing $\sqrt{x^{2}}=x$ without discussing the sign is the most widespread mistake on oblique asymptotes, and it gives a wrong line on one side of the graph out of two. Numerical check: for $x=10^{6}$, $f(x)-x\approx 1.4999989$; for $x=-10^{6}$, $f(x)+x\approx -1.5000011$.

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b) The Fundamental Theorem gives $F'=f$: the sign of $f$ drives the variations of $F$. Now $f>0$ on $(0, 4)$ and $f<0$ on $(4, 8)$. So $F$ INCREASES on $[0, 4]$ and DECREASES on $[4, 8]$, with a maximum at $x=4$, of value $F(4)=8$. The student's confusion is reading the graph as if it were $F$: what is drawn is $f$, that is, the DERIVATIVE of $F$. At $x=2$, $f$ reaches its maximum, which means $F$ is increasing FASTEST there, not that it is highest. $F$ peaks where its derivative is zero and changes sign, so at $x=4$, exactly when $f$ crosses the axis from positive to negative.

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b) The area cannot be zero: the curve clearly moves away from the axis. You must split at the ZEROS of $f$, that is where $x(x-2)(x+2)=0$, so at $-2$, $0$ and $2$. On $\left(-2, 0\right)$, $f>0$ (for example $f(-1)=3$); on $\left(0, 2\right)$, $f<0$ (for example $f(1)=-3$). So $\mathcal{A}=\int_{-2}^{0}f(x)\,dx-\int_{0}^{2}f(x)\,dx$. The first piece equals $\left[\frac{x^{4}}{4}-2x^{2}\right]_{-2}^{0}=0-(4-8)=4$ and the second $\left[\frac{x^{4}}{4}-2x^{2}\right]_{0}^{2}=-4$. Hence $\mathcal{A}=4-(-4)=8$.

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b) We require $\frac{M(b-a)^{5}}{180n^{4}}<10^{-4}$ with $M=24$, that is $\frac{24}{180n^{4}}<10^{-4}$, so $n^{4}>\frac{24\times 10^{4}}{180}=1333.33$, so $n>6.04$. The next integer would be 7, but $n$ must be EVEN for Simpson: so take $n=8$. Check: $\frac{24}{180\times 4096}=3.26\times 10^{-5}<10^{-4}$ ✓. Note in passing that $n=6$ would not be enough, since $\frac{24}{180\times 1296}=1.03\times 10^{-4}$, very slightly above the threshold: rounding up to the next even integer is not an unnecessary precaution, here it is the difference between guaranteeing and not guaranteeing.

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a) Over one slice, the curve moves forward by $dx$ and up by $dy=f'(x)dx$: its length is the hypotenuse $\sqrt{dx^{2}+dy^{2}}=\sqrt{1+[f'(x)]^{2}}\,dx$, which you add up. Here $f'(x)=\frac{3}{2}x^{1/2}$: $L=\int_{0}^{4}\sqrt{1+\frac{9x}{4}}\,dx=\left[\frac{8}{27}\left(1+\frac{9x}{4}\right)^{3/2}\right]_{0}^{4}=\frac{8}{27}\left(10^{3/2}-1\right)\approx 9.07$: a little more than the chord $\sqrt{4^{2}+8^{2}}\approx 8.94$, as it should be.

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a) $x'(t)=2t$ and $y'(t)=3t^{2}-3$, so $\dfrac{dy}{dx}=\dfrac{3t^{2}-3}{2t}$ for $t\neq 0$. The quotient makes no sense at $t=0$, precisely where the tangent line will be vertical.

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a) $r=\sqrt{1+3}=2$. The point is in the SECOND quadrant ($x<0$, $y>0$). The calculator gives $\arctan\left(\frac{\sqrt{3}}{-1}\right)=\arctan\left(-\sqrt{3}\right)=-\frac{\pi}{3}$, which designates a point in the fourth quadrant: that is wrong. You must add $\pi$: $\theta=-\frac{\pi}{3}+\pi=\frac{2\pi}{3}$. Check: $2\cos\frac{2\pi}{3}=-1$ and $2\sin\frac{2\pi}{3}=\sqrt{3}$.

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b) The reasoning is wrong because the NUMBER of terms grows while each term shrinks. Each term $\frac{1}{n+k}$ is of order $\frac{1}{n}$, but there are $n$ of them: the total stays of order $n\cdot\frac{1}{n}=1$, so nothing tends to 0. It is the same illusion as for the harmonic series: infinitely many tiny contributions can add up to a nonzero amount, and “taking the limit term by term” is only allowed if the number of terms is FIXED. Correct calculation: $\sum_{k=1}^{n}\frac{1}{n+k}=\sum_{k=1}^{n}\frac{1}{n\left(1+\frac{k}{n}\right)}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}$, which is the Riemann sum of $f(x)=\frac{1}{1+x}$ on $[0,1]$. The limit is therefore $\int_{0}^{1}\frac{dx}{1+x}=\left[\ln(1+x)\right]_{0}^{1}=\ln 2\approx 0.693$. Numerical check: the sum is $0.6688$ for $n=10$ and $0.69314$ for $n=10^{5}$, the convergence is slow and from below.

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b) $(-1)^{n}\frac{n}{n+1}$: the general term does not tend to 0 (it oscillates between values close to $+1$ and $-1$): DIVERGENT by the divergence test: Leibniz does not apply ($b_{n}\to 1\ne 0$) and should not even be invoked. $\frac{n^{2}}{n^{4}+1}$: positive, compares with $\frac{n^{2}}{n^{4}}=\frac{1}{n^{2}}$: $\frac{n^{2}}{n^{4}+1}\le\frac{1}{n^{2}}$, p-series with $p=2$: CONVERGENT (absolutely, since it is positive).

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a) The sequence $\frac{1}{n^{2}}$ is positive, decreasing and tends to 0: the Alternating Series Test applies and the series converges. The remainder after $N$ terms satisfies $|R_{N}|\leq\dfrac{1}{(N+1)^{2}}$. We want $\dfrac{1}{(N+1)^{2}}<10^{-3}$, that is $(N+1)^{2}>1000$, so $N+1>31.6$ and $N\geq 31$. Check: with 31 terms, the actual gap with $\frac{\pi^{2}}{12}\approx 0.8224670$ is $5.0\times 10^{-4}$, indeed below the bound.

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b) $\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$. Since the shift is 2, the terms cancel two indices apart, and TWO terms REMAIN at the start: $S_{N}=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{N+1}-\frac{1}{N+2}\right)\longrightarrow\frac{1}{2}\cdot\frac{3}{2}=\frac{3}{4}$. The general rule: a shift of $k$ leaves $k$ terms on each side.

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a) Ratio Test: $\left|\frac{x^{n+1}}{n+1}\cdot\frac{n}{x^{n}}\right|=|x|\frac{n}{n+1}\to|x|$: convergence for $|x|<1$, so $R=1$. Endpoints: at $x=1$, the harmonic series $\sum\frac{1}{n}$ DIVERGES; at $x=-1$, the alternating harmonic series CONVERGES (Alternating Series Test). Interval: $[-1, 1)$: closed on one side, open on the other, and it took both tests to know it.

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a) The trap is in the exponent: $e^{2x+y}=e^{2x}\,e^{y}$, not $e^{2x}+e^{y}$. Factor out $e^{y}$: $\dfrac{dy}{dx}=e^{y}\left(e^{2x}+1\right)$, a SEPARABLE equation. $e^{-y}\,dy=\left(e^{2x}+1\right)dx$, so $-e^{-y}=\dfrac{e^{2x}}{2}+x+C$. At the point $(0, 0)$: $-1=\frac{1}{2}+C$, $C=-\frac{3}{2}$. Hence $e^{-y}=\dfrac{3}{2}-x-\dfrac{e^{2x}}{2}$ and $y=-\ln\left(\dfrac{3}{2}-x-\dfrac{e^{2x}}{2}\right)$. At $x=-\frac{1}{2}$: $\frac{3}{2}+\frac{1}{2}-\frac{e^{-1}}{2}=2-\frac{1}{2e}\approx 1.816$, so $y\approx -0.597$. The solution exists only as long as the argument of the logarithm is positive: $3-2x-e^{2x}>0$. This function is strictly decreasing, equals $1$ at $0$ and vanishes between $0.39$ and $0.40$; by bisection, the right endpoint is about $0.396$. As you approach it, $e^{-y}\to 0$ and $y\to+\infty$: the solution blows up in finite time, although the right-hand side looked harmless.

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