Integration Practice: Methods, Area and Volumes

Use $\Delta t=2/n$, $t_k=2k/n$, and $\sum_{k=1}^n k=n(n+1)/2$. A rectangle has units $(\mathrm{m/s})(\mathrm{s})$.

Let $n$ be a positive integer. Each of the $n$ equal subintervals of $[0,2]$ has width $\Delta t=2/n$, measured in seconds. The index $k=1,\ldots,n$ labels these subintervals from left to right: the $k$th is $[2(k-1)/n,2k/n]$, with right endpoint $t_k=2k/n$. The product $v(t_k)\Delta t$ estimates the displacement during that subinterval. Let $S_n$ denote the total estimated displacement, in metres; the summation sign adds these $n$ contributions.

The right-endpoint sum is $$\begin{gathered}S_n=\sum_{k=1}^n v(t_k)\Delta t\\=\sum_{k=1}^n\left(1+\frac{4k}{n}\right)\frac2n.\end{gathered}$$ The factor $\Delta t$ is essential: velocity values alone do not add up to a displacement.

To evaluate the sum, use $\sum_{k=1}^n 1=n$ and $\sum_{k=1}^n k=n(n+1)/2$.

Split the sum and simplify: $$S_n=2+\frac8{n^2}\frac{n(n+1)}2=6+\frac4n.$$ Thus $S_4=7\,\mathrm m$. The four velocities are 2, 3, 4 and 5 m/s, each multiplied by 0.5 s.

The limit is $$\lim_{n\to\infty}S_n=6\,\mathrm m=\int_0^2(1+2t)\,dt.$$ Velocity is increasing, so each right-endpoint rectangle overestimates its interval’s area. Velocity is positive throughout, so displacement and distance are both 6 m.

Check the result

The integral also equals the trapezoid area $\tfrac12(1+5)(2)=6$. The estimate must exceed this value.

Common errors

• Omitting the width of each rectangle.
• Using n+1 sample points for n rectangles.
• Calling every velocity integral a distance without checking the velocity sign.

The upper endpoint is $g(x)=x^2$. The fundamental theorem gives the integrand at $g(x)$, multiplied by $g^{\prime}(x)$.

At $x=1$, $$A(1)=\left[t+\frac{t^4}{4}\right]_0^1=\frac54.$$

The integrand $1+t^3$ is continuous on $[0,4]$, which contains every integration interval because $0\le x^2\le4$. Apply the fundamental theorem and the chain rule: $$\begin{gathered}A^{\prime}(x)=(1+(x^2)^3)(2x)\\=2x(1+x^6).\end{gathered}$$ Here $t$ is a dummy integration variable, while $x$ controls the endpoint. Both replacing $t$ by $x$ and omitting the factor $2x$ lose this dependence.

An independent calculation gives $A(x)=x^2+x^8/4$. Differentiating yields $2x+2x^7$, the same answer.

Check the result

For $0<x<2$, the endpoint increases and the integrand is positive, so $A^{\prime}(x)>0$.

Common errors

• Writing $1+x^3$ instead of evaluating at $x^2$.
• Writing $1+x^6$ without multiplying by $2x$.

The derivative of the denominator appears in the numerator. Try $u=1+x^2$.

Let $u=1+x^2$, so $du=2x\,dx$. The endpoints become $u(0)=1$ and $u(1)=2$.

Then $$I=\int_1^2\frac{du}{u}=[\ln u]_1^2=\ln2.$$ Since $u>0$, the logarithm is defined throughout the transformed interval.

Alternatively, an antiderivative in the original variable is $\ln(1+x^2)$; its derivative is the original integrand. Use x-endpoints only after returning to x.

Check the result

On $[1,2]$, $1/2\le1/u\le1$, so $1/2\le I\le1$. The value $\ln2\approx0.6931$ passes.

Common errors

• Keeping 0 and 1 as the bounds of an integral in u.
• Dropping the factor 2, or leaving an unexplained x after the change of variable.

Differentiating x simplifies it. Integrating $e^{2x}$ keeps the same functional form but introduces a factor $1/2$.

Choose $u=x$ and $dv=e^{2x}dx$. Then $du=dx$ and $v=e^{2x}/2$.

Integration by parts gives $$\begin{gathered}J=\left[\frac{xe^{2x}}2\right]_0^1-\frac12\int_0^1e^{2x}\,dx\\=\frac{e^2}2-\frac{e^2-1}4\\=\frac{e^2+1}4.\end{gathered}$$

The remaining integral is simpler because the polynomial factor has disappeared. An antiderivative is $e^{2x}(2x-1)/4$.

Check the result

Differentiating $e^{2x}(2x-1)/4$ gives $xe^{2x}$. Also $x\le xe^{2x}\le e^2x$ on $[0,1]$, so $1/2\le J\le e^2/2$; the answer $\approx2.0973$ fits.

Common errors

• Using $v=e^{2x}$ rather than $e^{2x}/2$.
• Losing the minus sign in integration by parts.
• Forgetting the lower-endpoint contribution of the antiderivative.

Factor the denominator as $(x+1)(x+2)$. Write the fraction as $a/(x+1)+b/(x+2)$.

The numerator has lower degree than the denominator, so polynomial division is unnecessary. The denominator factors as $x^2+3x+2=(x+1)(x+2)$, a product of two distinct linear factors. Partial fractions are useful because they express the integrand as a sum of simpler fractions whose antiderivatives are logarithms. Let $a$ and $b$ be real constants to be determined. For $x\ne-1$ and $x\ne-2$, write $$\begin{gathered}\frac{x+3}{x^2+3x+2}\\=\frac{a}{x+1}+\frac{b}{x+2}.\end{gathered}$$ Multiplying the proposed decomposition by $(x+1)(x+2)$ gives $$x+3=a(x+2)+b(x+1).$$ Comparing coefficients yields $a+b=1$ and $2a+b=3$, so $a=2$, $b=-1$.

Recombine to check: $$\begin{gathered}\frac2{x+1}-\frac1{x+2}\\=\frac{2(x+2)-(x+1)}{(x+1)(x+2)}\\=\frac{x+3}{x^2+3x+2}.\end{gathered}$$

Both factors are positive on $[0,1]$. Therefore $$\begin{gathered}K=[2\ln(x+1)-\ln(x+2)]_0^1\\=3\ln2-\ln3\\=\ln\frac83.\end{gathered}$$ On any interval where the denominator is nonzero, the primitive uses $2\ln|x+1|-\ln|x+2|+C$.

Check the result

The integrand is positive on the interval; $K=\ln(8/3)\approx0.9808>0$. Differentiate the logarithmic primitive for a second check.

Common errors

• Assigning the coefficients without multiplying back.
• Using the antiderivative at x=1 without subtracting its value at x=0.
• Forgetting that logarithm formulas must respect the interval’s domain.

(a) Keep one $\sin t\,dt$ and replace the remaining $\sin^2t$ using an identity. (b) A substitution $x=3\sin\theta$ turns $9-x^2$ into $9\cos^2\theta$. Specify a range on which the square root has a known sign.

(a) A trigonometric integrand is already present. Write $\sin^3t\cos^2t=(1-\cos^2t)\cos^2t\sin t$. Let $u=\cos t$, $du=-\sin t\,dt$. Then $$\begin{gathered}\int\sin^3t\cos^2t\,dt\\=-\int(u^2-u^4)du\\=-\frac{\cos^3t}{3}+\frac{\cos^5t}{5}+C.\end{gathered}$$

(b) Start with an algebraic radical. Choosing $u=9-x^2$ does not directly absorb the factor $x^2dx$; a trigonometric substitution handles the radical systematically. Set $$\begin{gathered}x=3\sin\theta,\\0<\theta<\frac\pi2,\\dx=3\cos\theta\,d\theta.\end{gathered}$$ This range corresponds exactly to $0<x<3$ and makes $\cos\theta>0$. Thus $\sqrt{9-x^2}=3\cos\theta$, not an unspecified signed cosine.

After substituting every factor, $$\begin{gathered}\int\frac{x^2}{\sqrt{9-x^2}}dx\\=9\int\sin^2\theta\,d\theta\\=\frac92\theta-\frac94\sin(2\theta)+C.\end{gathered}$$ The last step uses $\sin^2\theta=(1-\cos2\theta)/2$.

The notation $\arcsin$ means the inverse sine with values in $[-\pi/2,\pi/2]$. To return to x, use $$\begin{gathered}\theta=\arcsin(x/3),\\\sin\theta=x/3,\\\cos\theta=\sqrt{9-x^2}/3.\end{gathered}$$ Hence $\sin2\theta=2x\sqrt{9-x^2}/9$, and the answer is $$\frac92\arcsin(x/3)-\frac{x\sqrt{9-x^2}}2+C.$$ The right triangle in the figure encodes these ratios; it applies to the stated positive-x interval.

Check the result

For (a), differentiating gives $\sin t\cos^2t(1-\cos^2t)=\sin^3t\cos^2t$.

For (b), let $r=\sqrt{9-x^2}>0$. The derivative is $$\frac9{2r}-\frac r2+\frac{x^2}{2r}=\frac{x^2}{r},$$ which recovers the integrand.

Common errors

• In (a), losing the minus sign from $du=-\sin t\,dt$.
• In (b), forgetting the 3 in $dx$, or the resulting factor 9.
• Replacing $\sqrt{\cos^2\theta}$ by $\cos\theta$ without checking the angle range.
• Leaving a mixture of x and θ in the final answer.
• Using a triangle side itself where a trigonometric ratio is required.

Solve $x^2=x$. Test a point between the intersections or factor $x-x^2$ to determine its sign.

Intersections satisfy $x(x-1)=0$, so $x=0$ and $x=1$. On $0<x<1$, $x-x^2=x(1-x)>0$, so $y=x$ lies above $y=x^2$.

A vertical strip has height $x-x^2$. Its area accumulates as $$\begin{gathered}B=\int_0^1(x-x^2)\,dx\\=\left[\frac{x^2}2-\frac{x^3}3\right]_0^1\\=\frac16.\end{gathered}$$ The area is $1/6$ square coordinate units.

Check the result

The strip height lies between 0 and 1/4, and the interval has width 1; therefore $0<B<1/4$. The value $1/6$ fits.

Common errors

• Dividing by x while solving for intersections and losing x=0.
• Subtracting the upper curve from the lower curve.
• Using an interval from the diagram without solving the intersection equations.

A radius is a distance from $y=-1$. Both boundaries are above that axis. Subtract the two disk areas, not the radii before squaring.

Use vertical slices with $0\le x\le1$. The upper curve $y=x$ is farther from the axis: $$R(x)=x-(-1)=x+1.$$ The lower curve gives $$r(x)=x^2-(-1)=x^2+1.$$ A gap separates the region from the axis, so the section has a central hole.

The cross-sectional area is $$\begin{gathered}C(x)=\pi\{R(x)^2-r(x)^2\}\\=\pi\{(x+1)^2-(x^2+1)^2\}.\end{gathered}$$ Therefore $$\begin{gathered}V=\pi\int_0^1(2x-x^2-x^4)\,dx\\=\pi\left[x^2-\frac{x^3}3-\frac{x^5}5\right]_0^1\\=\frac{7\pi}{15}.\end{gathered}$$ The answer is in cubic coordinate units.

The region’s vertical height $x-x^2$ is $R-r$. Squaring that height would give $(R-r)^2$, which is not $R^2-r^2$; it loses the information about how far the slice lies from the rotation axis.

Check the result

On $[0,1]$, $R\ge r>0$, so every cross-sectional area is nonnegative. Also $$R^2-r^2=(x-x^2)(x+x^2+2).$$ Since the second factor lies between 2 and 4, Question 7 implies $\pi/3\le V\le2\pi/3$. The value $7\pi/15$ lies between these bounds.

Common errors

• Measuring radii from y=0 instead of y=-1.
• Using a disk and filling the central hole.
• Squaring the difference of radii rather than subtracting their squares.
• Integrating with respect to y while using vertical slices and x-bounds.