Integrals

(a) Write out the terms. The sum runs over $k=3,4,5,6,7$: that is $7-3+1=5$ terms, not $4$. They are $9-6=3$, $16-8=8$, $25-10=15$, $36-12=24$, $49-14=35$, and $3+8+15+24+35=85$.

By linearity. The power-sum formulas start at $1$, so subtract the two missing terms:

$$\sum _{k=3}^7{k}^2=\sum _{k=1}^7{k}^2-\left(1^2+2^2\right)=\frac{7\cdot 8\cdot 15}{6}-5=140-5=135,\sum _{k=3}^{7}k=\frac{7\cdot 8}{2}-3=25.$$

Then ${\sum }_{k=3}^7(k^2-2k)=135-2\cdot 25=85$, agreeing with the direct count.

(b) Expand, then split by linearity. $(2i-1{)}^2=4i^2-4i+1$, so

$$\sum _{i=1}^n(2i-1{)}^2=4\sum _{i=1}^n{i}^2-4\sum _{i=1}^{n}i+\sum _{i=1}^{n}1=4\cdot \frac{n(n+1)(2n+1)}{6}-4\cdot \frac{n(n+1)}{2}+n.$$

The last term is $n$, not $1$: a constant summed $n$ times is counted $n$ times. Factor out $\frac{n}{3}$:

$$\frac{n}{3}[2(n+1)(2n+1)-6(n+1)+3]=\frac{n}{3}[4n^2+6n+2-6n-6+3]=\frac{n\left(4n^2-1\right)}{3}=\frac{n(2n-1)(2n+1)}{3}.$$

Check at $n=3$: $1+9+25=35$, and $\frac{3\cdot 5\cdot 7}{3}=35$. (At $n=1$ the formula gives $\frac{1\cdot 1\cdot 3}{3}=1$, also correct.)

(c) Divide by $n^3$ and let $n\to \infty$.

$$\frac{1}{n^3}\cdot \frac{n(2n-1)(2n+1)}{3}=\frac{1}{3}\left(2-\frac{1}{n}\right)\left(2+\frac{1}{n}\right)⟶\frac{1}{3}\cdot 2\cdot 2=\frac{4}{3}.$$

To see the integral, rewrite the summand: $\frac{(2i-1{)}^2}{n^3}={\left(\frac{2i-1}{n}\right)}^2\cdot \frac{1}{n}=4{\left(\frac{2i-1}{2n}\right)}^2\cdot \frac{1}{n}$. With $\Delta x=\frac{1}{n}$ on $[0,1]$, the point $\frac{2i-1}{2n}$ is the midpoint of the $i$th subinterval, so the sum is the midpoint Riemann sum $M_n$ for ${\int }_0^{1}4x^{2}dx$, whose value is $4\cdot \frac{1}{3}=\frac{4}{3}$, as found.

(a) Set up the partition. $\Delta x=\frac{2-0}{4}=\frac{1}{2}$, with partition points $0,\frac{1}{2},1,\frac{3}{2},2$ and function values $4,\frac{15}{4},3,\frac{7}{4},0$.

Left endpoints are the first four points, $0,\frac{1}{2},1,\frac{3}{2}$:

$$L_4=\frac{1}{2}\left(4+\frac{15}{4}+3+\frac{7}{4}\right)=\frac{1}{2}\cdot \frac{25}{2}=\frac{25}{4}.$$

Right endpoints are the last four, $\frac{1}{2},1,\frac{3}{2},2$:

$$R_4=\frac{1}{2}\left(\frac{15}{4}+3+\frac{7}{4}+0\right)=\frac{1}{2}\cdot \frac{17}{2}=\frac{17}{4}.$$

Midpoints are $\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4}$, with values $4-\frac{1}{16}=\frac{63}{16}$, $4-\frac{9}{16}=\frac{55}{16}$, $4-\frac{25}{16}=\frac{39}{16}$, $4-\frac{49}{16}=\frac{15}{16}$. Their sum is $\frac{172}{16}=\frac{43}{4}$, so

$$M_4=\frac{1}{2}\cdot \frac{43}{4}=\frac{43}{8}.$$

Each sum has exactly four terms; using all five partition points would double-count a strip.

(b) Monotonicity decides it. $f^{\prime }(x)=-2x\le 0$ on $[0,2]$, so $f$ is decreasing. On each subinterval the largest value of $f$ is at the left endpoint and the smallest at the right, so every left rectangle contains its strip of the region and every right rectangle sits inside it. Hence

$$R_4\le A\le L_4,\text{that is}\frac{17}{4}\le A\le \frac{25}{4}.$$

The gap is $L_4-R_4=\frac{25}{4}-\frac{17}{4}=2$, and the single-strip formula gives $[f(0)-f(2)]\Delta x=4\cdot \frac{1}{2}=2$: all the rectangles except the outermost ones cancel in the difference.

(c) The limit of $R_n$. With $n$ subintervals, $\Delta x=\frac{2}{n}$ and $x_i=\frac{2i}{n}$, so $f(x_i)=4-\frac{4i^2}{n^2}$ and

$$R_n=\sum _{i=1}^n\left(4-\frac{4i^2}{n^2}\right)\frac{2}{n}=\frac{8}{n}\cdot n-\frac{8}{n^3}\sum _{i=1}^n{i}^2=8-\frac{8}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}=8-\frac{4}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right).$$

As $n\to \infty$ the parentheses tend to $1$ and $2$, so $A=8-\frac{8}{3}=\frac{16}{3}$.

Sandwich: $\frac{17}{4}=\frac{51}{12}\le \frac{64}{12}\le \frac{75}{12}=\frac{25}{4}$. The errors are $L_4-A=\frac{11}{12}$, $A-R_4=\frac{13}{12}$ and $M_4-A=\frac{129}{24}-\frac{128}{24}=\frac{1}{24}$. The midpoint sum is far closer than either endpoint sum, and it is a slight overestimate; the graph of $4-x^2$ is concave down, and the numerical-integration rules explain why midpoints overshoot on such a curve.

(a) Features of the drawing. The graph is a V with vertex $(0,-2)$, slope $-1$ to the left of $0$ and slope $+1$ to the right. Zeros: $|x|=2$, so $x=-2$ and $x=2$. Endpoint values: $f(-3)=1$, $f(4)=2$, and the vertex $f(0)=-2$.

So the graph runs from $(-3,1)$ down through $(-2,0)$, dips below the axis to $(0,-2)$, climbs back through $(2,0)$ and up to $(4,2)$. Three triangles are shaded: a small one above the axis on $[-3,-2]$, a large one below the axis on $[-2,2]$ with its apex at the vertex, and a medium one above on $[2,4]$. Use two different shadings for above and below.

(b) Triangles. Left triangle: legs $1$ and $1$, area $\frac{1}{2}$. Middle triangle: base $4$ (from $-2$ to $2$), height $2$, area $\frac{1}{2}\cdot 4\cdot 2=4$. Right triangle: legs $2$ and $2$, area $2$.

Net area counts the region below the axis negatively:

$${\int }_{-3}^{4}f(x)dx=\frac{1}{2}-4+2=-\frac{3}{2}.$$

Total area counts every region positively:

$${\int }_{-3}^4|f(x)|dx=\frac{1}{2}+4+2=\frac{13}{2}.$$

They differ because $f$ changes sign on the interval, so the integral cancels the part below the axis against the parts above, while the total area adds them. Check of the net value by linearity: ${\int }_{-3}^4|x|dx-{\int }_{-3}^{4}2dx=\left(\frac{9}{2}+8\right)-14=-\frac{3}{2}$.

(c) Reflect in the axis. $2-|x|=-f(x)$, so its graph is the graph of $f$ reflected across the $x$-axis: the same three triangles, with the middle one now above and the outer two below. By linearity with the constant $-1$,

$${\int }_{-3}^4\left(2-|x|\right)dx=-{\int }_{-3}^{4}f(x)dx=\frac{3}{2},$$

while the total area is unchanged at $\frac{13}{2}$, since $|-f|=|f|$.

(a) Velocity first. Velocity is an antiderivative of acceleration: $v(t)=6t^2-6t+C$. The condition $v(1)=3$ gives $6-6+C=3$, so $C=3$ and $v(t)=6t^2-6t+3$.

Then position. $s(t)=2t^3-3t^2+3t+D$. The condition $s(2)=5$ gives $16-12+6+D=10+D=5$, so $D=-5$:

$$s(t)=2t^3-3t^2+3t-5.$$

Check: $s^{\prime }(t)=6t^2-6t+3$, so $s^{\prime }(1)=3$; $s^{\prime \prime }(t)=12t-6=a(t)$; and $s(2)=16-12+6-5=5$. The two conditions are given at different times, but each still removes exactly one constant, provided the velocity constant is fixed before the second antidifferentiation is done.

(b) On $(0,\infty )$. There $|x|=x$, so $\int \frac{1}{x}dx=\ln x+C$, and $g(1)=\ln 1+C=C=2$ gives $g(x)=\ln x+2$. Check: $g^{\prime }(x)=\frac{1}{x}$ and $g(1)=2$.

The two-piece function. For $x>0$, $G^{\prime }(x)=\frac{1}{x}$ directly. For $x<0$, $\ln |x|=\ln (-x)$, whose derivative is $\frac{-1}{-x}=\frac{1}{x}$; the constants differentiate to $0$. So $G^{\prime }(x)=\frac{1}{x}$ at every point of the domain $(-\infty ,0)\cup (0,\infty )$, and $G$ is an antiderivative of $\frac{1}{x}$ there.

Yet $G(x)-\left(\ln |x|+2\right)$ equals $0$ for $x>0$ and $5$ for $x<0$: not a single constant. There is no contradiction, because the theorem is a statement about an interval. Its proof uses the corollary of the Mean Value Theorem that a function with zero derivative on an interval is constant, and that corollary needs the interval to connect any two points. The domain of $\frac{1}{x}$ is two disjoint intervals, and on each one separately $G$ is $\ln |x|$ plus a constant, exactly as the theorem says. The formula $\int \frac{1}{x}dx=\ln |x|+C$ really carries one constant per interval.

(a) The bounds property. On $[0,1]$ we have $0\le x^3\le 1$, so $1\le 1+x^3\le 2$, and since the square root is increasing, $1\le \sqrt{1+x^3}\le \sqrt{2}$. The integrand is continuous, so the integral exists by the integrability theorem, and the bounds property with $m=1$, $M=\sqrt{2}$ and $b-a=1$ gives

$$1\cdot 1\le I\le \sqrt{2}\cdot 1.$$

(b) The inequality. For $x\ge 0$ both sides are nonnegative, so it is enough to compare squares:

$${\left(1+\frac{x^3}{2}\right)}^2=1+x^3+\frac{x^6}{4}\ge 1+x^3,$$

and taking the (increasing) square root of both sides gives $\sqrt{1+x^3}\le 1+\frac{x^3}{2}$, with equality only at $x=0$.

The comparison property now applies to the two continuous functions on $[0,1]$:

$$I\le {\int }_0^1\left(1+\frac{x^3}{2}\right)dx={\left[x+\frac{x^4}{8}\right]}_0^1=1+\frac{1}{8}=\frac{9}{8},$$

the right-hand integral being evaluated by linearity and Part 2 of the Fundamental Theorem. So $1\le I\le \frac{9}{8}$. (The true value is about $1.111$.)

(c) Proof. Suppose, for contradiction, that $f(c)>0$ for some $c$ in $[a,b]$. Apply continuity at $c$ with $\epsilon =\frac{f(c)}{2}$: there is $\delta >0$ such that $|f(x)-f(c)|<\frac{f(c)}{2}$, hence $f(x)>\frac{f(c)}{2}$, for every $x$ in $[a,b]$ with $|x-c|<\delta$.

Put $p=\max \left(a,c-\frac{\delta }{2}\right)$ and $q=\min \left(b,c+\frac{\delta }{2}\right)$. Then $[p,q]\subseteq [a,b]$, its length $\ell =q-p$ is positive (even if $c$ is an endpoint, the interval has room on at least one side), and $f>\frac{f(c)}{2}$ on all of $[p,q]$.

By additivity,

$${\int }_a^{b}f={\int }_a^{p}f+{\int }_p^{q}f+{\int }_q^{b}f.$$

The first and third integrals are $\ge 0$ by positivity, since $f\ge 0$ (and they are $0$ if the interval is degenerate). By the bounds property with lower bound $m=\frac{f(c)}{2}$ on $[p,q]$, the middle one satisfies ${\int }_p^{q}f\ge \frac{f(c)}{2}\ell >0$. Hence ${\int }_a^{b}f>0$, contradicting the hypothesis. So $f(c)\le 0$ for every $c$, and together with $f\ge 0$ this forces $f\equiv 0$. $■$

The example. Let $f(x)=0$ for $0\le x<1$ and $f(1)=1$. This is bounded with a single jump, so it is integrable. Any Riemann sum $\sum f(x_i^{\ast })\Delta x$ has at most one nonzero term, the last, and only if the sample point $x_n^{\ast }$ is exactly $1$; the sum is therefore either $0$ or $\Delta x=\frac{1}{n}$, and both tend to $0$. So ${\int }_0^{1}f=0$ and $f\ge 0$, yet $f(1)=1\ne 0$. The proof breaks at the very first step: without continuity there is no interval of positive length on which $f$ stays away from $0$, and a single point has no width for the integral to see.

(a) Expand and divide through. For $x>0$, $(\sqrt{x}-1{)}^2=x-2\sqrt{x}+1$, so

$$\frac{(\sqrt{x}-1{)}^2}{x}=1-2x^{-1/2}+x^{-1}.$$

The first two are power-rule entries; the third is the one power the power rule cannot handle, and it is the logarithm:

$$\int \left(1-2x^{-1/2}+x^{-1}\right)dx=x-2\cdot \frac{x^{1/2}}{1/2}+\ln |x|+C=x-4\sqrt{x}+\ln x+C,$$

where $|x|=x$ because the integrand lives on $x>0$. Check: $\frac{d}{dx}\left(x-4\sqrt{x}+\ln x\right)=1-\frac{2}{\sqrt{x}}+\frac{1}{x}$.

(b) Split the fraction into table factors.

$$\frac{\sin \theta }{{\cos }^2\theta }=\frac{1}{\cos \theta }\cdot \frac{\sin \theta }{\cos \theta }=\sec \theta \tan \theta ,$$

which is a table entry: the integral is $\sec \theta +C$. Check: $\frac{d}{d\theta }\sec \theta =\sec \theta \tan \theta$.

(c) Divide out the polynomial part. Since $2x^2+3=2(x^2+1)+1$,

$$\frac{2x^2+3}{x^2+1}=2+\frac{1}{1+x^2},\text{so}\int \frac{2x^2+3}{x^2+1}dx=2x+\arctan x+C.$$

Check: $\frac{d}{dx}\left(2x+\arctan x\right)=2+\frac{1}{1+x^2}$.

(d) Divide by $e^x$. $\frac{e^{2x}-1}{e^x}=e^x-e^{-x}$. An antiderivative of $e^{-x}$ is $-e^{-x}$, since $\frac{d}{dx}\left(-e^{-x}\right)=e^{-x}$. Hence

$$\int \left(e^x-e^{-x}\right)dx=e^x+e^{-x}+C.$$

Alternatively, $e^x-e^{-x}=2\sinh x$ by definition of $\sinh$, and the table gives $\int 2\sinh xdx=2\cosh x+C$. The two agree because $2\cosh x=e^x+e^{-x}$. Check: $\frac{d}{dx}\left(e^x+e^{-x}\right)=e^x-e^{-x}$.

(a) Compute the integral, then solve for $c$. $f(x)=\sqrt{x}$ is continuous on $[0,4]$, so the theorem applies. By Part 2 of the Fundamental Theorem,

$${\int }_0^4\sqrt{x}dx={\left[\frac{2}{3}{x}^{3/2}\right]}_0^4=\frac{2}{3}\cdot 8=\frac{16}{3}.$$

The theorem asks for $c$ with $\sqrt{c}(4-0)=\frac{16}{3}$, so $\sqrt{c}=\frac{4}{3}$ and $c=\frac{16}{9}$. This lies in $[0,4]$, and it is the only solution because $\sqrt{x}$ is one-to-one. The rectangle has base $[0,4]$ and height $f\left(\frac{16}{9}\right)=\frac{4}{3}$, so its area is $\frac{16}{3}$, the same as the region under the curve; the curve passes through the top edge of the rectangle at $x=\frac{16}{9}$. The height $\frac{4}{3}$ is the average value of $\sqrt{x}$ on $[0,4]$.

(b) Proof. Let $g(x)=f(x)-x$. It is continuous on $[0,1]$ as the difference of two continuous functions, and by linearity

$${\int }_0^{1}g(x)dx={\int }_0^{1}f(x)dx-{\int }_0^{1}xdx=\frac{1}{2}-\frac{1}{2}=0.$$

The Mean Value Theorem for integrals (hypothesis: $g$ continuous on the closed interval $[0,1]$) supplies $c$ in $[0,1]$ with ${\int }_0^{1}g=g(c)(1-0)$, that is $g(c)=0$, that is $f(c)=c$. $■$

The number $\frac{1}{2}$ is exactly ${\int }_0^{1}xdx$: the hypothesis says $f$ and the identity function have the same average on $[0,1]$, and two continuous functions with the same average must cross.

(c) A jump defeats the theorem. Take $f(x)=0$ for $0\le x<1$ and $f(x)=1$ for $1\le x\le 2$. It is bounded with one jump, hence integrable, and by additivity and geometry ${\int }_0^{2}f=0+1=1$ (the second piece is a unit square). The theorem would need $f(c)\cdot 2=1$, so $f(c)=\frac{1}{2}$, but $f$ takes only the values $0$ and $1$. No such $c$ exists.

The proof uses continuity twice. First, the Extreme Value Theorem, to know that the minimum $m$ and maximum $M$ are values actually taken by $f$. Second, and decisively, the Intermediate Value Theorem, to pass from \"the average $\frac{1}{b-a}{\int }_a^{b}f$ lies between two values of $f$\" to \"the average is a value of $f$\". Here the average $\frac{1}{2}$ lies between the values $0$ and $1$, and the jump lets $f$ step over it.

(a) The graph of $f$. A horizontal segment at height $2$ on $[0,2]$; a straight segment from $(2,2)$ down to $(4,-2)$, crossing the axis at $t=3$; a horizontal segment at height $-2$ on $[4,6]$. The pieces join up, since $6-2\cdot 2=2$ and $6-2\cdot 4=-2$, so $f$ is continuous.

Values of $g$ by area. $g(0)=0$. On $[0,1]$ a $1\times 2$ rectangle: $g(1)=2$. On $[0,2]$ a $2\times 2$ rectangle: $g(2)=4$. On $[2,3]$ a triangle above the axis with base $1$ and height $2$, area $1$: $g(3)=5$. On $[3,4]$ the mirror-image triangle below the axis, area $1$, counted negatively: $g(4)=4$. On $[4,5]$ and $[5,6]$ rectangles of area $2$ below the axis: $g(5)=2$ and $g(6)=0$.

(b) Read $g$ off $f$. Since $f$ is continuous on $[0,6]$, Part 1 of the Fundamental Theorem gives $g^{\prime }(x)=f(x)$. Now $f>0$ on $[0,3)$ and $f<0$ on $(3,6]$, so $g$ is increasing on $[0,3]$, decreasing on $[3,6]$, and has its maximum at $x=3$, where $g(3)=5$.

Concavity comes from $g^{\prime \prime }=f^{\prime }$ wherever $f^{\prime }$ exists: $f^{\prime }=0$ on $(0,2)$ and on $(4,6)$, so $g$ is a straight line on each of those (slopes $2$ and $-2$); and $f^{\prime }=-2$ on $(2,4)$, so $g$ is concave down there. Explicitly, $g(x)=2x$ on $[0,2]$; on $[2,4]$,

$$g(x)=4+{\int }_2^x(6-2t)dt=4+{\left[6t-t^2\right]}_2^x=6x-x^2-4,$$

a parabola with vertex at $x=3$ and $g(3)=5$; and on $[4,6]$, $g(x)=4-2(x-4)=12-2x$. The formulas agree at the joins: $6\cdot 2-4-4=4$ and $6\cdot 4-16-4=4$.

(c) The graph of $g$. It starts at the origin and rises as a straight line of slope $2$ to $(2,4)$; continues as a downward-opening parabolic arc through the maximum $(3,5)$ to $(4,4)$; then descends as a straight line of slope $-2$ through $(5,2)$ to $(6,0)$. Draw the parabola tangent to the incoming line at $(2,4)$ and to the outgoing line at $(4,4)$.

There is no corner because the slope of $g$ is $f$, and $f$ is continuous. At $x=2$ the line has slope $2$ and the parabola has slope $6-2\cdot 2=2$; at $x=4$ the parabola has slope $6-2\cdot 4=-2$ and the line has slope $-2$. A corner in $g$ would require a jump in $f$; a change of formula is not a jump.

Total area. ${\int }_0^6|f(t)|dt=4+1+1+4=10$. On the sketch of $f$ this is the sum of the two rectangles and the two triangles, regardless of side. On the sketch of $g$ it is the total climb plus the total descent: $g$ rises from $0$ to $5$ and then falls from $5$ back to $0$, and $5+5=10$. The net area, by contrast, is the final height $g(6)=0$.

(a) Displacement is $\int v$; distance is $\int |v|$. By the net change theorem,

$$\text{displacement}={\int }_0^4\left(3t^2-12t+9\right)dt={\left[t^3-6t^2+9t\right]}_0^4=64-96+36=4\text{m}.$$

For distance, find where $v$ changes sign: $v(t)=3\left(t^2-4t+3\right)=3(t-1)(t-3)$, which is positive on $[0,1)$, negative on $(1,3)$ and positive on $(3,4]$. With $F(t)=t^3-6t^2+9t$ we have $F(0)=0$, $F(1)=4$, $F(3)=27-54+27=0$ and $F(4)=4$, so the three pieces are

$${\int }_0^{1}v=4,{\int }_1^{3}v=-4,{\int }_3^{4}v=4.$$

Distance $={\int }_0^4|v(t)|dt=4+4+4=12$ m. Check: the signed pieces add to $4-4+4=4$, the displacement. The particle goes forward $4$ m, back $4$ m, forward $4$ m, and ends $4$ m from where it started having travelled $12$ m.

(b) Volume is $50$ plus the integral of the rate. Let $V(t)=50+{\int }_0^{t}r(s)ds$. By Part 1 of the Fundamental Theorem, $V^{\prime }(t)=r(t)$, so $V$ is largest where $r$ changes from positive to negative. Now $r(t)=0$ when $3\sqrt{t}=12$, that is $t=16$; $r>0$ on $[0,16)$ and $r<0$ on $(16,25]$. So the tank is fullest at $t=16$ minutes, and

$$V(16)=50+{\int }_0^{16}\left(12-3\sqrt{s}\right)ds=50+{\left[12s-2s^{3/2}\right]}_0^{16}=50+(192-128)=114\text{litres}.$$

At $t=25$,

$$V(25)=50+{\left[12s-2s^{3/2}\right]}_0^{25}=50+(300-250)=100\text{litres}.$$

Check by additivity: ${\int }_{16}^{25}r=50-64=-14$, and $114-14=100$. The question asked for volumes, which are integrals of $r$ itself; the total amount of water that passed through the inlet, $64+14=78$ litres, would be the integral of $|r|$.

(a) A $u$-antiderivative evaluated at $x$-limits. The substitution and the antiderivative $-\frac{1}{3}{u}^{3/2}$ are both correct. The wrong step is \"evaluating from $0$ to $2$\": those are the $x$-limits, and the expression is a function of $u$. The theorem on substitution in definite integrals says the limits must be converted to $u$-values, $x=0\mapsto u=4$ and $x=2\mapsto u=0$, or else one must return to $x$ first. Mixing the two is the violation.

The answer could have been rejected on sight: the integrand $x\sqrt{4-x^2}$ is $\ge 0$ on $[0,2]$, so by positivity the integral cannot be negative.

Correct evaluation, changing the limits:

$${\int }_0^{2}x\sqrt{4-x^2}dx=-\frac{1}{2}{\int }_4^0{u}^{1/2}du=\frac{1}{2}{\int }_0^4{u}^{1/2}du=\frac{1}{2}\cdot \frac{2}{3}{\left[u^{3/2}\right]}_0^4=\frac{1}{3}\cdot 8=\frac{8}{3}.$$

Or back-substitute: ${\left[-\frac{1}{3}{\left(4-x^2\right)}^{3/2}\right]}_0^2=-\frac{1}{3}(0-8)=\frac{8}{3}$. Check by differentiating: $\frac{d}{dx}\left[-\frac{1}{3}(4-x^2{)}^{3/2}\right]=-\frac{1}{3}\cdot \frac{3}{2}(4-x^2{)}^{1/2}\cdot (-2x)=x\sqrt{4-x^2}$.

(b) Symmetry of the interval is only half the hypothesis. The theorem on even and odd functions needs a symmetric interval and an integrand of definite parity, and only an odd integrand gives $0$. Here $x^3+x^2$ is neither even nor odd: at $x=1$ it is $2$, at $x=-1$ it is $0$. The sentence \"integrals over symmetric intervals cancel\" is simply false; ${\int }_{-2}^{2}1dx=4$.

Correct evaluation, splitting by linearity into the odd and even parts:

$${\int }_{-2}^2{x}^{3}dx=0(\text{odd}),{\int }_{-2}^2{x}^{2}dx=2{\int }_0^2{x}^{2}dx=2\cdot \frac{8}{3}=\frac{16}{3}(\text{even}),$$

so the integral is $\frac{16}{3}$. Direct check: ${\left[\frac{x^4}{4}+\frac{x^3}{3}\right]}_{-2}^2=\left(4+\frac{8}{3}\right)-\left(4-\frac{8}{3}\right)=\frac{16}{3}$.

(c) There is no product rule for integrals. The number $0$ is right, but the step $\int fg=\int f\cdot \int g$ is not a rule at all: linearity covers sums and constant multiples, nothing more. A counterexample on a different interval shows the step is unreliable: on $[0,\frac{\pi }{2}]$, $\int \sin xdx=\int \cos xdx=1$, while ${\int }_0^{\pi /2}\sin x\cos xdx={\left[\frac{1}{2}{\sin }^{2}x\right]}_0^{\pi /2}=\frac{1}{2}\ne 1\cdot 1$.

Correct evaluation: substitute $u=\sin x$, $du=\cos xdx$; the limits become $\sin 0=0$ and $\sin \pi =0$, so

$${\int }_0^{\pi }\sin x\cos xdx={\int }_0^{0}udu=0$$

by the convention ${\int }_a^a=0$. Equivalently ${\left[\frac{1}{2}{\sin }^{2}x\right]}_0^{\pi }=0-0=0$. The value is $0$ for a reason the bad argument never touched: $\sin x\cos x=\frac{1}{2}\sin 2x$ is positive on $(0,\frac{\pi }{2})$ and negative on $(\frac{\pi }{2},\pi )$, with the two humps congruent, so the net area cancels. The total area between the curve and the axis is $\frac{1}{2}+\frac{1}{2}=1$.

(a) First proof, by substitution. Fix $x$ and write $g(-x)={\int }_0^{-x}f(t)dt$. Substitute $t=-u$, so $dt=-du$; the limits $t=0$ and $t=-x$ become $u=0$ and $u=x$. The theorem on substitution in definite integrals applies, since the inner function $u\mapsto -u$ has continuous derivative and $f$ is continuous everywhere:

$$g(-x)={\int }_0^{x}f(-u)(-1)du=-{\int }_0^{x}f(-u)du=-{\int }_0^{x}f(u)du=-g(x),$$

using $f(-u)=f(u)$ in the last step. So $g(-x)=-g(x)$ for every $x$: $g$ is odd. $■$

Second proof, by differentiation. Let $h(x)=g(x)+g(-x)$. By Part 1 of the Fundamental Theorem, $g^{\prime }(x)=f(x)$ for all $x$, and by the chain rule $\frac{d}{dx}g(-x)=-g^{\prime }(-x)=-f(-x)=-f(x)$, since $f$ is even. Therefore

$$h^{\prime }(x)=f(x)-f(x)=0\text{for every }x.$$

A function with zero derivative on an interval is constant (the corollary of the Mean Value Theorem), and the interval here is all of $\mathbb{R}$. Since $h(0)=g(0)+g(0)=0$ (an integral from $0$ to $0$ is $0$), the constant is $0$, so $g(-x)=-g(x)$. $■$ The pair $f=\cos$, $g=\sin$ is the model case.

(b) Proof. Define $\phi (a)={\int }_a^{a+p}f(t)dt$. By additivity, $\phi (a)={\int }_0^{a+p}f-{\int }_0^{a}f=g(a+p)-g(a)$. Both terms are differentiable by Part 1 of the Fundamental Theorem together with the chain rule (the corollary on variable limits), and

$${\phi }^{\prime }(a)=f(a+p)\cdot 1-f(a)=f(a)-f(a)=0$$

by periodicity. So ${\phi }^{\prime }$ vanishes on the whole real line, and by the Mean Value Theorem corollary $\phi$ is constant: ${\int }_a^{a+p}f={\int }_0^{p}f$ for every $a$. $■$ In words, a window of one full period captures the same amount wherever it is placed.

(c) Apply (b) with $p=\pi$. The function $|\sin t|$ is continuous and has period $\pi$, since $|\sin (t+\pi )|=|-\sin t|=|\sin t|$. By additivity, split $[1,1+3\pi ]$ into the three windows $[1,1+\pi ]$, $[1+\pi ,1+2\pi ]$ and $[1+2\pi ,1+3\pi ]$, each of length one period. By (b) each contributes ${\int }_0^{\pi }|\sin t|dt$, and on $[0,\pi ]$ the sine is nonnegative, so

$${\int }_0^{\pi }|\sin t|dt={\int }_0^{\pi }\sin tdt={\left[-\cos t\right]}_0^{\pi }=1+1=2.$$

Hence ${\int }_1^{1+3\pi }|\sin t|dt=3\cdot 2=6$.

Cross-check the hard way. The sign changes of $\sin$ inside $[1,1+3\pi ]$ are at $\pi$, $2\pi$ and $3\pi$. The pieces are ${\int }_1^{\pi }\sin tdt=1+\cos 1$, then $2$ on $[\pi ,2\pi ]$ and $2$ on $[2\pi ,3\pi ]$, and on $[3\pi ,3\pi +1]$ the sine is negative, giving ${\int }_{3\pi }^{3\pi +1}(-\sin t)dt={\left[\cos t\right]}_{3\pi }^{3\pi +1}=-\cos 1-(-1)=1-\cos 1$. The total is $(1+\cos 1)+2+2+(1-\cos 1)=6$: the $\cos 1$ terms cancel, exactly as (b) predicts they must.

The exact value. $I={\left[\frac{x^4}{4}\right]}_0^2=4$.

(a) Trapezoidal and midpoint. $\Delta x=\frac{1}{2}$, endpoints $0,\frac{1}{2},1,\frac{3}{2},2$ with cubes $0,\frac{1}{8},1,\frac{27}{8},8$:

$$T_4=\frac{1/2}{2}\left[0+2\left(\frac{1}{8}+1+\frac{27}{8}\right)+8\right]=\frac{1}{4}\left[2\cdot \frac{36}{8}+8\right]=\frac{1}{4}\cdot 17=\frac{17}{4}.$$

Midpoints $\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4}$ have cubes $\frac{1}{64},\frac{27}{64},\frac{125}{64},\frac{343}{64}$, which sum to $\frac{496}{64}=\frac{31}{4}$, so