Corrected exercises: definite and improper integrals, 201-SN3-RE (201-NYB)

$\int_{2}^{\infty}\frac{dx}{x^{3}}=\lim\limits_{b\to\infty}\left[-\frac{1}{2x^{2}}\right]_{2}^{b}=\lim\limits_{b\to\infty}\left(-\frac{1}{2b^{2}}+\frac{1}{8}\right)=\frac{1}{8}$. Converges, as the $p$-test predicts with $p=3>1$.

$\int_{-\infty}^{0}e^{3x}\,dx=\lim\limits_{a\to-\infty}\left[\frac{e^{3x}}{3}\right]_{a}^{0}=\lim\limits_{a\to-\infty}\left(\frac{1}{3}-\frac{e^{3a}}{3}\right)=\frac{1}{3}$, since $e^{3a}\to 0$ as $a\to-\infty$. Pitfall: forgetting the factor $\frac{1}{3}$ in the antiderivative.

The function is unbounded near $0$. $\int_{0}^{1}x^{-1/3}\,dx=\lim\limits_{a\to 0^{+}}\left[\frac{3}{2}x^{2/3}\right]_{a}^{1}=\frac{3}{2}-\lim\limits_{a\to 0^{+}}\frac{3}{2}a^{2/3}=\frac{3}{2}$. Converges: at the endpoint $0$, $p=\frac{1}{3}<1$. Pitfall: the antiderivative of $x^{-1/3}$ divides by $\frac{2}{3}$, so it multiplies by $\frac{3}{2}$.

The singularity is at the endpoint $4$. $\int_{0}^{4}(4-x)^{-1/2}\,dx=\lim\limits_{b\to 4^{-}}\left[-2\sqrt{4-x}\right]_{0}^{b}=\lim\limits_{b\to 4^{-}}\left(-2\sqrt{4-b}\right)+2\sqrt{4}=0+4=4$. Pitfall: the minus sign in the antiderivative, which comes from the derivative $-1$ of $4-x$.

The singularity is at the endpoint $2$. $\int_{2}^{3}\frac{dx}{(x-2)^{2}}=\lim\limits_{a\to 2^{+}}\left[-\frac{1}{x-2}\right]_{a}^{3}=\lim\limits_{a\to 2^{+}}\left(-1+\frac{1}{a-2}\right)=+\infty$: the integral diverges to $+\infty$. This is the $p$-test at a finite endpoint, with $p=2\ge 1$. Pitfall: evaluating the antiderivative at $2$ as if nothing were wrong.

Substitution $u=x^{2}$, $du=2x\,dx$: an antiderivative is $-\frac{1}{2}e^{-x^{2}}$. $\int_{0}^{\infty}xe^{-x^{2}}\,dx=\lim\limits_{b\to\infty}\left[-\frac{1}{2}e^{-x^{2}}\right]_{0}^{b}=\lim\limits_{b\to\infty}\left(-\frac{1}{2}e^{-b^{2}}\right)+\frac{1}{2}=\frac{1}{2}$.

Two infinite limits: split at $0$. An antiderivative is $\frac{1}{2}\arctan\frac{x}{2}$. $\int_{0}^{\infty}\frac{dx}{x^{2}+4}=\lim\limits_{b\to\infty}\frac{1}{2}\arctan\frac{b}{2}=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}$, and likewise $\int_{-\infty}^{0}\frac{dx}{x^{2}+4}=\frac{\pi}{4}$. Both halves converge, so the integral equals $\frac{\pi}{2}\approx 1.571$. Pitfall: the antiderivative $\arctan x$, which only fits $\frac{1}{x^{2}+1}$ and would give $\pi$.

Substitution $u=\ln x$: an antiderivative is $\frac{(\ln x)^{2}}{2}$. $\int_{1}^{\infty}\frac{\ln x}{x}\,dx=\lim\limits_{b\to\infty}\frac{(\ln b)^{2}}{2}=+\infty$: the integral diverges to $+\infty$. Comparison already said so: $\frac{\ln x}{x}\ge\frac{1}{x}$ as soon as $x\ge e$.

Substitution $u=\ln x$, $du=\frac{dx}{x}$: $\int\frac{dx}{x(\ln x)^{2}}=-\frac{1}{\ln x}$. $\int_{e}^{\infty}\frac{dx}{x(\ln x)^{2}}=\lim\limits_{b\to\infty}\left(-\frac{1}{\ln b}\right)+\frac{1}{\ln e}=0+1=1$. Converges, whereas $\int_{e}^{\infty}\frac{dx}{x\ln x}$ diverges: the square changes everything.

The function is unbounded at $0$, INSIDE $[-1, 1]$: split there. An antiderivative is $3\sqrt[3]{x}$. $\int_{0}^{1}=\lim\limits_{a\to 0^{+}}\left(3-3\sqrt[3]{a}\right)=3$ and $\int_{-1}^{0}=\lim\limits_{b\to 0^{-}}3\sqrt[3]{b}-3\sqrt[3]{-1}=0+3=3$. Both pieces converge, so the integral equals $6$. Pitfall: computing $\left[3\sqrt[3]{x}\right]_{-1}^{1}$ without splitting; here the number comes out right by luck, but the same mistake on $\frac{1}{x^{2}}$ gives a negative area.

On the right, the heights are $f(1),f(2),f(3),f(4)=1, 4, 9, 16$: $S_{R}=(1+4+9+16)(1)=30$. On the left: $f(0),f(1),f(2),f(3)$: $S_{L}=(0+1+4+9)(1)=14$. Since $x^{2}$ is INCREASING on $[0, 4]$, each left rectangle stays under the curve and each right rectangle sticks out above it: $14\le\int_{0}^{4}x^{2}dx\le 30$. The bound is crude ($\Delta x$ is too large), but it is certain.

With $\Delta x=\frac{4}{n}$ and $x_{i}=\frac{4i}{n}$: $S_{n}=\sum_{i=1}^{n}\left(\frac{4i}{n}\right)^{2}\frac{4}{n}=\frac{64}{n^{3}}\sum_{i=1}^{n}i^{2}=\frac{64}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{64}{6}\cdot\frac{(n+1)(2n+1)}{n^{2}}$. As $n\to\infty$, $\frac{(n+1)(2n+1)}{n^{2}}\to 2$: $S_{n}\to\frac{64\times 2}{6}=\frac{64}{3}\approx 21.33$. The exact value does fall between 14 and 30.

Fundamental Theorem (evaluation part): if $F'=f$ on $[a,b]$, then $\int_{a}^{b}f=F(b)-F(a)$. Here $F(x)=\frac{x^{3}}{3}$: $\int_{0}^{4}x^{2}dx=\frac{64}{3}-0=\frac{64}{3}$. All the work of the limit in b) is packed once and for all into the theorem: it is THE bridge between derivative and area, and the very reason antiderivatives exist.

Other direction: $g'(x)=\cos(x^{2})$, directly, without looking for an antiderivative. For $h$, the limit is composite: chain rule, $h'(x)=\cos((x^{2})^{2})\cdot 2x=2x\cos(x^{4})$. Yet $\cos(t^{2})$ has NO antiderivative expressible with the usual functions (a Fresnel integral): the theorem still lets you differentiate, study the variations and sketch $g$. An integral with a variable limit BUILDS new functions, and the theorem says how they vary.

Antiderivatives: $3\sqrt{x}=3x^{1/2}$ gives $2x^{3/2}$, and $\frac{1}{x^{2}}=x^{-2}$ gives $-\frac{1}{x}$. So $\left[2x^{3/2}-\frac{1}{x}\right]_{1}^{4}=\left(16-\frac{1}{4}\right)-(2-1)=15.75-1=\frac{59}{4}=14.75$.

$u=x^{2}$, $du=2x\,dx$: the $x\,dx$ on top is exactly $\frac{du}{2}$. Limits: $x=0\Rightarrow u=0$, $x=2\Rightarrow u=4$. $\int_{0}^{2}xe^{x^{2}}dx=\frac{1}{2}\int_{0}^{4}e^{u}du=\frac{e^{4}-1}{2}\approx 26.80$. Changing the limits together with the variable avoids going back to $x$: shorter, and fewer errors.

By parts: $\int u\,dv=uv-\int v\,du$. DIFFERENTIATE what simplifies and INTEGRATE what you know how to integrate: $u=\ln x$ (its derivative $\frac{1}{x}$ kills the logarithm), $dv=x\,dx$, $v=\frac{x^{2}}{2}$. $\int_{1}^{e}x\ln x\,dx=\left[\frac{x^{2}}{2}\ln x\right]_{1}^{e}-\int_{1}^{e}\frac{x}{2}dx=\frac{e^{2}}{2}-\left[\frac{x^{2}}{4}\right]_{1}^{e}=\frac{e^{2}}{2}-\frac{e^{2}-1}{4}=\frac{e^{2}+1}{4}\approx 2.10$.

$x^{3}\sqrt{x^{2}+1}$ is ODD (an odd function times an even one) and the interval $[-2, 2]$ is symmetric: the integral is 0, each area on the right being cancelled by its opposite on the left. No computation, and the antiderivative would have been painful. Average value: $\bar{f}=\frac{1}{\pi}\int_{0}^{\pi}\sin x\,dx=\frac{1}{\pi}[-\cos x]_{0}^{\pi}=\frac{2}{\pi}\approx 0.637$. Interpretation: the rectangle with base $\pi$ and height $\frac{2}{\pi}$ has exactly the same area as the arch of the sine curve; this is also why the rectified average voltage of a sinusoidal signal is $\frac{2}{\pi}$ of its amplitude.

$x^{2}=x+2\iff x^{2}-x-2=0\iff(x-2)(x+1)=0$: intersections at $(-1, 1)$ and $(2, 4)$, as in the figure. Between them, the line is on top: $A=\int_{-1}^{2}(x+2-x^{2})dx=\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{27}{6}=\frac{9}{2}=4.5$.

The curves CROSS at $x=\frac{\pi}{4}$ (where $\tan x=1$): you must split there, otherwise the two pieces cancel each other. $\int_{0}^{\pi/4}(\cos x-\sin x)dx=[\sin x+\cos x]_{0}^{\pi/4}=\sqrt{2}-1$, and by symmetry the second piece is also $\sqrt{2}-1$. Total area: $2\sqrt{2}-2\approx 0.83$. Reflex: always look for the crossings BEFORE integrating a difference.

Each slice perpendicular to the axis is a disk of radius $\sqrt{x}$: $V=\pi\int_{0}^{4}(\sqrt{x})^{2}dx=\pi\int_{0}^{4}x\,dx=\pi\cdot 8=8\pi\approx 25.1$. Squaring the radius removes the root: paraboloids are the most cooperative solids of the disk method.

$\int_{0}^{2\pi}\sin x\,dx=[-\cos x]_{0}^{2\pi}=0$: the positive arch and the negative arch cancel exactly, this is the SIGNED area. The GEOMETRIC area counts everything as positive: $\int_{0}^{2\pi}|\sin x|dx=2\int_{0}^{\pi}\sin x\,dx=4$. When the question says “area of the region”, it always means the geometric one: split at the zeros of the function and add absolute values. Mixing the two up is the most common area error of the course.

Additivity over intervals: $\int_{0}^{5}=\int_{0}^{3}+\int_{3}^{5}$, so $\int_{3}^{5}f=4-7=-3$. The negative sign is no surprise: on $[3, 5]$ the function clearly dips below the axis. Orientation: swapping the limits changes the sign, $\int_{5}^{0}f=-\int_{0}^{5}f=-4$. Equal limits: $\int_{3}^{3}f=0$, an interval of zero width accumulates nothing. Linearity: $\int_{0}^{5}(2f+1)=2\int_{0}^{5}f+\int_{0}^{5}1\,dx=2(4)+5=13$, the second term being the area of the rectangle of height 1 and base 5.

Monotonicity property: if $m\le f\le M$ on $[a,b]$, then $m(b-a)\le\int_{a}^{b}f\le M(b-a)$, because integration preserves inequalities. On $[0, 2]$, $x^{3}$ goes from 0 to 8, so $1\le 1+x^{3}\le 9$, and since the square root is increasing, $1\le\sqrt{1+x^{3}}\le 3$. Multiplying by the width $b-a=2$: $2\le\int_{0}^{2}\sqrt{1+x^{3}}\,dx\le 6$. The actual value is about $3.24$: the bound is correct, though wide. Nothing required an antiderivative, and that is the whole point: $\sqrt{1+x^{3}}$ has no elementary one, yet we already know where its integral lies.

From $-|f|\le f\le|f|$ and the monotonicity of the integral: $-\int_{a}^{b}|f|\le\int_{a}^{b}f\le\int_{a}^{b}|f|$, which is exactly $\left|\int f\right|\le\int|f|$. Strict case: $f=\sin$ on $[0, 2\pi]$ gives $\left|\int_{0}^{2\pi}\sin x\,dx\right|=0$ whereas $\int_{0}^{2\pi}|\sin x|dx=4$. Cancellation of signs is precisely what the absolute value forbids: $0<4$. Refuting the product: $\int_{0}^{1}x^{2}dx=\frac{1}{3}$, while $\left(\int_{0}^{1}x\,dx\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$. Since $\frac{1}{3}\ne\frac{1}{4}$, the integral of a product is NOT the product of the integrals. The integral is linear, not multiplicative, and that is exactly why integration by parts exists.

Mean Value Theorem for Integrals: if $f$ is CONTINUOUS on $[a,b]$, there exists $c\in[a,b]$ such that $f(c)=\frac{1}{b-a}\int_{a}^{b}f$; in other words the function reaches its own average value at least once. Here $\frac{1}{3}\int_{0}^{3}x^{2}dx=\frac{1}{3}\cdot 9=3$, so $c^{2}=3$ and $c=\sqrt{3}\approx 1.73$, well inside $[0, 3]$. It is NOT the midpoint: $f(1.5)=2.25\ne 3$. The average of a convex function is pulled upward by the fast growth on the right, so $c$ lies to the right of the midpoint. Without continuity, everything collapses: the step function equals 0 on $[0, 1)$ and 1 on $[1, 2]$, its integral over $[0, 2]$ is 1 and its average $\frac{1}{2}$, but it ONLY takes the values 0 and 1, never $\frac{1}{2}$. The Intermediate Value Theorem secretly powers the Mean Value Theorem for Integrals: remove continuity and nothing is left.

Each piece is a triangle. On $[0, 2]$: a triangle with base 2 and height 4, area $\frac{2\times 4}{2}=4$, so $F(2)=4$. On $[2, 4]$: the same triangle flipped, area 4, still above the axis, so $F(4)=4+4=8$. On $[4, 6]$: a triangle BELOW the axis, base 2, height 2, area 2 counted negatively: $F(6)=8-2=6$. On $[6, 8]$: another $-2$, so $F(8)=4$. We do have $F(8)<F(4)$ because $F$ does not accumulate quantity, it accumulates SIGNED area: after $x=4$, $f$ is negative and each new slice SUBTRACTS area from the total. An account whose deposit rate turns negative keeps existing; it empties.

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.

$F''=f'$, that is, the SLOPE of the given graph. The slopes are successively $+2$ on $[0, 2]$, $-2$ on $[2, 4]$, $-1$ on $[4, 6]$, $+1$ on $[6, 8]$. So $F$ is concave up on $[0, 2]$, concave down on $[2, 6]$, concave up on $[6, 8]$. The inflection points are at $x=2$ and $x=6$, where the slope of $f$ CHANGES SIGN. At $x=4$, the slope goes from $-2$ to $-1$: it changes value but stays negative, so $F''$ stays negative and the concavity does not change. No inflection point at 4. The trap is a classic: what counts is the sign of $f'$, never the mere fact that a slope changes. Note how the roles are shared: the zeros of $f$ give the extrema of $F$, the extrema of $f$ give the inflection points of $F$.

$G(0)=\int_{2}^{0}f=-\int_{0}^{2}f=-4$. $G(8)=\int_{2}^{8}f=F(8)-F(2)=4-4=0$. In general $G(x)=F(x)-F(2)=F(x)-4$: the two functions differ by a CONSTANT, and their graphs are vertical translates of each other by 4 units. That is exactly what the constant of integration means: changing the lower limit of an integral with a variable upper limit only changes the reference level, the point where the accumulation is declared to be zero. Since $G'=F'=f$, the two functions have the same variations, the same concavity, the same maximum at $x=4$ and the same inflection points at 2 and 6. A vertical translation moves no critical point: that is why any antiderivative is always enough to answer a question about variations.