Unit 02 - Essential problems

Arc length

01

Arc length - reversed $x$ and $y$ roles

Find the arc length of the curve that satisfies the equation $x=\frac{1}{12}{y}^3+y^{-1}$ over $y\in [\mathrm{1,2}]$.

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Solution

Solutions - 0080-01

(1) Integral formula for arclength:

$$S={\int }_c^d\sqrt{1+{(g^{\prime })}^2}dx$$

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(2) Work out integrand:

$$\begin{array}{c}g(x)=\frac{1}{12}{y}^3+y^{-1}\gg \gg g^{\prime }(x)=\frac{1}{4}{y}^2-y^{-2}\\ \\ {(g^{\prime })}^2\gg \gg \frac{1}{16}{y}^4-\frac{1}{2}+y^{-4}\\ \\ 1+{(g^{\prime })}^2\gg \gg \frac{1}{16}{y}^4+\frac{1}{2}+y^{-4}\gg \gg {(\frac{1}{4}{y}^2+y^{-2})}^2\\ \\ \sqrt{1+{(g^{\prime })}^2}\gg \gg \frac{1}{4}{y}^2+y^{-2}\end{array}$$

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(3) Integrate:

$$\begin{array}{c}S\gg \gg {\int }_1^2\frac{1}{4}{y}^2+y^{-2}dy\\ \\ \gg \gg {\frac{1}{12}{y}^3-y^{-1}|}_1^2\gg \gg \frac{13}{12}\end{array}$$Link to original

Surface areas of revolutions - thin bands

01

Surface area: revolved cubic

The curve $y=x^3$ over $x\in [\mathrm{0,2}]$ is revolved around the $x$-axis.

Find the area of the resulting surface.

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Solution

Solutions - 0090-01

(1) Integral formula for surface area, revolution about $x$-axis:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+{\left(f^{\prime }(x)\right)}^2}dx$$

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(2) Work out integrand:

$$a=0,b=2,f(x)=x^3,f^{\prime }(x)=3x^2$$

Then:

$$\sqrt{1+{(f^{\prime })}^2}\gg \gg \sqrt{1+9x^4}$$

So:

$$S\gg \gg {\int }_0^{2}2\pi x^3\sqrt{1+9x^4}dx$$

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(3) Perform $u$-sub with $u=1+9x^4$ and so $du=36x^{3}dx$:

$${\int }_0^{2}2\pi x^3\sqrt{1+9x^4}dx\gg \gg \frac{\pi }{18}{\int }_1^{145}\sqrt{u}du$$

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(4) Integrate:

$$\begin{array}{cc}\frac{\pi }{18}{\int }_1^{145}\sqrt{u}du& \gg \gg \frac{\pi }{27}{\left[u^{3/2}\right]}_1^{145}\\ & \\ & \gg \gg \frac{\pi }{27}(145\sqrt{145}-1)\end{array}$$Link to original

03

Surface area: parabolic reflector

A parabolic reflector is given by rotating the curve $y=x^2$ around the $y$-axis for $x\in [\mathrm{0,2}]$.

What is the surface area of this reflector?

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Solution

Solutions - 0090-03

Method 1: integrate in $x$

(1) Integral formula for surface area:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+{(f^{\prime })}^2}dx\gg \gg 2\pi {\int }_0^{2}x\sqrt{1+4x^2}dx$$

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(2) Integrate: perform $u$-sub with $u=1+4x^2$ and so $du=8xdx$:

$$\begin{array}{c}S\gg \gg \frac{\pi }{4}{\int }_1^{17}\sqrt{u}du\\ \\ \gg \gg \frac{\pi }{4}{\frac{2}{3}{u}^{\frac{3}{2}}|}_1^{17}\\ \\ \gg \gg \frac{\pi }{6}({17}^{3/2}-1)\end{array}$$

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Method 2: integrate in $y$

(1) Integral formula for surface area using $g(y)=\sqrt{y}$:

$$\begin{array}{c}S={\int }_c^{d}2\pi g(y)\sqrt{1+{(g^{\prime })}^2}dy\gg \gg {\int }_0^{4}2\pi \sqrt{y}\sqrt{1+\frac{1}{4}{y}^{-1}}dy\\ \\ \gg \gg \pi {\int }_0^4\sqrt{4y+1}dy\end{array}$$

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(2) Integrate: perform $u$-sub with $u=4y+1$ and so $du=4dy$:

$$\begin{array}{c}\pi {\int }_0^4\sqrt{4y+1}dy\gg \gg \frac{\pi }{4}{\int }_1^{17}\sqrt{u}du\\ \\ \gg \gg \frac{\pi }{4}{\frac{2}{3}{u}^{\frac{3}{2}}|}_1^{17}\\ \\ \gg \gg \frac{\pi }{6}({17}^{3/2}-1)\end{array}$$Link to original

Hydrostatic pressure

03

Fluid force on triangular plates

For diagrams 1, 2, 3 (L to R) below, set up an integral to compute the hydrostatic force on the plate.

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Solution

Solutions - 0100-03

(1) Integral formula:

$$F={\int }_a^b\rho gh(y)w(y)dy$$

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Option 1:

(2) Using $y=0$ at water line, increasing downwards: (a) Left:

$$F={\int }_0^4\rho gy(3+\frac{0-3}{4}y)dy$$

(b) Center:

$$F={\int }_2^5\rho gy(0+\frac{2-0}{3}(y-2))dy$$

(c) Right:

$$F={\int }_0^4\rho gy(4+\frac{0-4}{5}(y+1))dy$$

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Option 2:

(2) Using $y=0$ at top of shape, increasing downwards: (a) Left:

$$F={\int }_0^4\rho gy(3+\frac{0-3}{4}y)dy$$

(b) Center:

$$F={\int }_0^3\rho g(y+2)(0+\frac{2-0}{3}y)dy$$

(c) Right:

$$F={\int }_1^5\rho g(y-1)(4+\frac{0-4}{5}y)dy$$Link to original

04

Fluid force on trapezoidal plates

For diagrams 1, 2, 3 (L to R) below, set up an integral to compute the hydrostatic force on the plate.

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Solution

Solutions - 0100-04

(a) Left:

Set $y=0$ at the water line, increasing downwards.

$$\begin{array}{c}w(y)=4+\frac{5-4}{3}(y-1)\gg \gg \frac{11}{3}+\frac{1}{3}y\\ \\ \gg \gg {\int }_1^4\rho gy(\frac{11}{3}+\frac{1}{3}y)dy\end{array}$$

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Alternative: Set $y=0$ at the top of the trapezoid. Obtain:

$${\int }_0^3\rho g(y+1)(4+\frac{5-4}{3}y)dy$$

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(b) Center:

Set $y=0$ at the water line, increasing downwards.

$$\begin{array}{c}w(y)=5+\frac{4-5}{3}(y+1)\gg \gg \frac{14}{3}-\frac{1}{3}y\\ \\ \gg \gg {\int }_0^2\rho gy(\frac{14}{3}-\frac{1}{3}y)dy\end{array}$$

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Alternative: Set $y=0$ at the top of the trapezoid. Obtain:

$${\int }_1^3\rho g(y-1)(5+\frac{4-5}{3}y)dy$$

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(c) Right:

Set $y=0$ at the water line, increasing downwards.

$$\begin{array}{cccc}& w_S(y)& =2& \text{(square)}\\ & w_T(y)& =2+\frac{0-2}{3}(y-2)& \text{(triangle)}\end{array}$$ $$\gg \gg {\int }_0^2\rho gy2dy+{\int }_2^5\rho gy(2-\frac{2}{3}(y-2))dy$$Link to original

05

Fluid force on circular plates

For diagrams 1, 2, 3 (L to R) below, set up an integral to compute the hydrostatic force on the plate.

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Solution

Solutions - 0100-05

(1) Integral formula:

$$F={\int }_a^b\rho gh(y)w(y)dy$$

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Option 1:

(2) Using $y=0$ at water line, increasing downwards: (a) Left:

$$F={\int }_1^5\rho gy\left(2\sqrt{2^2-{(y-3)}^2}\right)dy$$

(b) Center:

$$F={\int }_0^3\rho gy\left(2\sqrt{3^2-y^2}\right)dy$$

(c) Right:

$$F={\int }_0^2\rho gy\left(2\sqrt{4^2-{(2-y)}^2}\right)dy$$

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Option 2:

(2) Using $y=0$ at center of shape, increasing downwards: (a) Left:

$$F={\int }_{-2}^2\rho g(y+3)2\sqrt{2^2-y^2}dy$$

(b) Center:

$$F={\int }_0^3\rho gy\left(2\sqrt{3^2-y^2}\right)dy$$

(c) Right:

$$F={\int }_{-2}^0\rho g(y+2)2\sqrt{4^2-y^2}dy$$Link to original

Moments and CoM

01

CoM of a house

A “house” is the region bounded by the (non-regular) pentagon with vertex points at $(\mathrm{0,5})$, $(\mathrm{2,3})$, $(\mathrm{2,0})$, $(-\mathrm{2,0})$, $(-\mathrm{2,3})$. Find the CoM of the house using additivity of moments.

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Solution

Solutions - 0110-01

(1) Compute masses:

$$\begin{array}{c}{m}^{\text{total}}=m^{\text{rect}}+m^{\text{tri}}\\ \\ \gg \gg \rho A^{\text{rect}}+\rho A^{\text{tri}}\\ \\ \gg \gg \rho (4)(3)+\rho \left(\frac{1}{2}\right)(4)(2)\\ \\ \gg \gg 12\rho +4\rho \gg \gg 16\rho \end{array}$$

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(2) Consider symmetries of rectangle:

$${\text{CoM}}^{\text{rect}}=({\bar{x}}^{\text{rect}},{\bar{y}}^{\text{rect}})=(0,3/2)$$

Therefore $M_y^{\text{rect}}=0$ and:

$${\bar{y}}^{\text{rect}}=\frac{M_x^{\text{rect}}}{m^{\text{rect}}}$$

Therefore:

$$\begin{array}{c}{M}_x^{\text{rect}}={\bar{y}}^{\text{rect}}\cdot m^{\text{rect}}\\ \\ \gg \gg (3/2)\cdot 12\rho \gg \gg 18\rho \end{array}$$

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(3) Consider symmetry of triangle:

$${\bar{x}}^{\text{tri}}=0\gg \gg M_y^{\text{tri}}=0$$

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(4) Compute $M_x^{\text{tri}}$ by integration:

$$w(y)=4+\frac{0-4}{2}(y-3)\gg \gg 10-2y$$ $$\begin{array}{c}{M}_x^{\text{tri}}={\int }_3^5\rho yw(y)dy\\ \\ \gg \gg \rho {\int }_3^{5}y(10-2y)dy\gg \gg 44\rho /3\end{array}$$

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(5) Optional step: infer ${\bar{y}}^{\text{tri}}$:

$${\bar{y}}^{\text{tri}}=\frac{M_x^{\text{tri}}}{m^{\text{tri}}}\gg \gg \frac{44\rho /3}{4\rho }\gg \gg 11/3$$

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(6) Additivity of moments:

$$\begin{array}{c}{M}_x^{\text{total}}=M_x^{\text{rect}}+M_x^{\text{tri}}\\ \\ \gg \gg 18\rho +44\rho /3\gg \gg 98\rho /3\end{array}$$ $$\begin{array}{c}{M}_y^{\text{total}}=M_y^{\text{rect}}+M_y^{\text{tri}}=0+0=0\end{array}$$

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(7) Compute CoM:

$$\bar{x}=0,\bar{y}=\frac{M_x^{\text{total}}}{m^{\text{total}}}$$ $$\gg \gg \bar{y}\gg \gg \frac{98\rho /3}{16\rho }\gg \gg 49/24$$

Thus:

$$\text{CoM}=(0,49/24)$$Link to original

02

CoM of region between curves

Find the CoM of the region between the graph of $y=e^x$ and the graph of $y=1$ over $x\in [\mathrm{0,1}]$.

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Solution

Solutions - 0110-02

Option 1: In terms of $x$.

(1) Mass of region:

$$\begin{array}{c}m=\rho {\int }_0^1{e}^x-1dx\\ \\ \gg \gg \rho (e^x-x){|}_0^1\gg \gg \rho (e-2)\end{array}$$

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(2) Integral formula for $M_y$:

$$\begin{array}{ccc}& M_y={\int }_a^b\rho x(f(x)-g(x))dx& \\ & & \\ & \gg \gg \rho {\int }_0^{1}x(e^x-1)dx& \\ & & \\ & \gg \gg \rho {(x{e}^x-e^x-\frac{x^2}{2})|}_0^1\gg \gg \frac{\rho }{2}& \text{(Int. by parts)}\end{array}$$

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(3) Integral formula for $M_x$:

$$\begin{array}{c}{M}_x={\int }_a^b\frac{\rho }{2}(f{(x)}^2-g{(x)}^2)dx\\ \\ \gg \gg \frac{\rho }{2}{\int }_0^1{e}^{2x}-1dx\\ \\ \gg \gg \frac{\rho }{2}{(\frac{1}{2}{e}^{2x}-x)|}_0^1\gg \gg \frac{\rho }{4}(e^2-3)\end{array}$$

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(4) Compute $(\bar{x},\bar{y})$:

$$\begin{array}{c}(\bar{x},\bar{y})=(\frac{M_y}{m},\frac{M_x}{m})\\ \\ \gg \gg (\frac{\rho /2}{\rho (e-2)},\frac{\frac{\rho }{4}(e^2-3)}{\rho (e-2)})\\ \\ \gg \gg (\frac{1}{2(e-2)},\frac{e^2-3}{4(e-2)})\end{array}$$

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Option 2: In terms of $y$.

(1) Mass:

$$\begin{array}{c}m/\rho \gg \gg {\int }_1^{e}1-\ln ydy\\ \\ \gg \gg y-(y\ln y-y){|}_1^e\\ \\ \gg \gg e-e+e-(1-(0-1))\\ \\ \gg \gg e-2\end{array}$$

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(2) $M_x$:

$$\begin{array}{ccc}& M_x/\rho \gg \gg {\int }_1^{e}y(1-\ln y)dy& \\ & & \\ & \gg \gg \frac{1}{4}(e^2-3)& \text{(Int. by parts)}\end{array}$$

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(3) $M_y$:

$$\begin{array}{ccc}& M_y/\rho \gg \gg {\int }_1^e\frac{1}{2}(1^2-(\ln y{)}^2)dy& \\ & & \\ & \gg \gg 1/2& \text{(}\text{I}\text{n}\text{t}\text{.}\text{b}\text{y}\text{p}\text{a}\text{r}\text{t}\text{s}\text{,}u=(\ln y{)}^2\text{)}\end{array}$$Link to original

Work performed

01

Pumping water from hemispherical tank

A hemispherical tank (radius $10\mathrm{m}$) is full of water. A pipe allows water to be pumped out, but requires pumping up $2\mathrm{m}$ above the top of the tank.

(a) Set up an integral that expresses the total work required to pump all the water out of the tank, assuming it is completely full.

(b) Now assume the tank start out full just to $6\mathrm{m}$. What does the integral become?

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Solution

Solutions - 0120-01

(a) (1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg \int h(y)\rho gdV\\ \\ \gg \gg \int \rho gh(y)A(y)dy\end{array}$$

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(2) Setup:

Coordinate system: set $y=0$ at the top of the tank, increasing downwards.

Horizontal slice of the tank: disk of radius $r$ at depth $y$, satisfies:

$$r^2+y^2={10}^2,A(y)=\pi r^2=\pi ({10}^2-y^2)$$

Distance pumped up (add $2$ for the spigot): $h(y)=y+2$

Thus:

$$W\gg \gg {\int }_0^{10}\rho g\pi (y+2)({10}^2-y^2)dy$$

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(b) (1) Change upper bound, top of water at $y=4$:

$$W\gg \gg {\int }_4^{10}\rho g\pi (y+2)({10}^2-y^2)dy$$Link to original

02

Building a conical tower

Set up an integral that expresses the work done (against gravity) to build a circular cone-shaped tower of height $4\mathrm{m}$ and base radius $1.2\mathrm{m}$ out of a material with mass density $600\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$.

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Solution

Solutions - 0120-02

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg \int h(y)\rho gdV\\ \\ \gg \gg \int \rho gh(y)A(y)dy\end{array}$$

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Option 1: (2) Setup:

Set $y=0$ at the bottom, increasing upwards.

Radius of the cone with a QLIF:

$$r(y)=1.2+\frac{0-1.2}{4}y\gg \gg 1.2-0.3y$$

Horizontal slice of the cone tower: disk of radius $r$ at height $y$, satisfies:

$$A(y)=\pi r^2=\pi {(1.2-0.3y)}^2$$

The slice at $y$ is raised a distance of $h(y)=y$.

Thus:

$$W\gg \gg {\int }_0^4\rho g\pi y{(1.2-0.3y)}^{2}dy$$

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Option 2: (2) Setup:

Set $y=0$ at the top of the cone, increasing downwards.

Now $h(y)=4-y$ is the distance from the ground up to the height of a slice indexed by $y$.

Radius function:

$$r(y)=0+\frac{1.2-0}{4}y\gg \gg 0.3y$$

Thus:

$$W\gg \gg {\int }_0^4\rho g\pi (4-y){(0.3y)}^{2}dy$$Link to original

03

Work to raise a leaky bucket

A bucket of water is raised by a chain to the top of a $50$-foot building. The water is leaking out, and the chain is getting lighter.

The bucket weighs $4\mathrm{lbs}$, the initial water weighs $30\mathrm{lbs}$, and the chain weighs $0.25\mathrm{l}\mathrm{b}/\mathrm{f}\mathrm{t}$, and the water is leaking at a rate of $0.3\mathrm{l}\mathrm{b}\mathrm{s}/\mathrm{s}\mathrm{e}\mathrm{c}$ as the bucket is lifted at a constant rate of $3\mathrm{f}\mathrm{t}/\mathrm{s}\mathrm{e}\mathrm{c}$.

What is the total work required to raise the bucket of water?

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Solution

Solutions - 0120-03

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg {\int }_a^{b}F(y)dy\end{array}$$

Let $y=0$ at the ground and increase going up.

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(2) Compute force:

The force on the rope (at the top) when the bucket is at height $y$ is:

$$F=F_{\text{bucket}}+F_{\text{water}}+F_{\text{chain}}$$

We know $F_{\text{bucket}}=4\mathrm{lb}$.

Water is leaking at ${\displaystyle \frac{0.3}{3}}\mathrm{l}\mathrm{b}/\mathrm{f}\mathrm{t}$. Therefore:

$$F_{\text{water}}=30-\frac{0.3}{3}y$$

The weight of chain remaining is:

$$F_{\text{chain}}=0.25\cdot (50-y)$$

Put together:

$$\begin{array}{c}F=4+30-\frac{0.3}{3}y+0.25\cdot (50-y)\\ \\ \gg \gg 46.5-0.35y\end{array}$$

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(3) Integrate:

$$W={\int }_0^{50}(46.5-0.35y)dy\gg \gg 1887.5\mathrm{f}\mathrm{t}\cdot \mathrm{l}\mathrm{b}$$Link to original

Improper integrals

01

Comparison test

Use the comparison test to determine whether the integral converges:

$${\int }_1^{\infty }\frac{dx}{x^4+\sqrt{x}}$$

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Solution

Solutions - 0130-01

(1) Find comparable integrand:

Higher power dominates for large $x$:

$$\frac{1}{x^4+\sqrt{x}}⟶\frac{1}{x^4}\text{as }x\to \infty$$

Therefore, compare to $1/x^4$.

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(2) Make comparison:

$$\frac{1}{x^4+\sqrt{x}}<\frac{1}{x^4}\text{for all }x\ge 1$$

And:

$${\int }_1^{\infty }\frac{1}{x^4}dx\text{converges}$$

because it is a $p$-integral with $p=4>1$.

By the Comparison Test, we conclude that:

$${\int }_1^{\infty }\frac{1}{x^4+\sqrt{x}}dx\text{converges}$$Link to original

05

Computing improper integrals

For each integral below, give the limit interpretation of improper integral and then compute the limit. Based on that result, state whether the integral converges. If it converges, what is its value?

(a) ${\displaystyle {\int }_0^1\ln xdx}$ $$ (b) ${\displaystyle {\int }_{-\infty }^{+\infty }x{e}^{-x^2}dx}$ $$ (c) ${\displaystyle {\int }_1^2\frac{1}{{(x-1)}^2}dx}$

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Solution

Solutions - 0130-05

(a)

(1) Definition of improper integral:

$${\int }_0^1\ln xdx=\underset{R\to 0^{+}}{\lim }{\int }_R^1\ln xdx$$

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(2) Antiderivative and limit:

$$\begin{array}{ccc}& \gg \gg \underset{R\to 0^{+}}{\lim }x\ln x-x{|}_R^1& \\ & & \\ & \gg \gg \underset{R\to 0^{+}}{\lim }0-1-(R\ln R-R)\gg \gg -1\text{(converges)}& \text{(Note A)}\end{array}$$

Note A: Use L’Hopital:

$$R\ln R=\frac{\ln R}{1/R}{\gg \gg }^{d/dR}\frac{1/R}{-1/R^2}=-R\to 0$$

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(b)

(1) Definition of improper integral:

$${\int }_{-\infty }^{+\infty }x{e}^{-x^2}dx=\underset{R\to +\infty }{\lim }{\int }_0^{R}x{e}^{-x^2}dx+\underset{R\to -\infty }{\lim }{\int }_R^{0}x{e}^{-x^2}dx$$

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(2) Antiderivative and limit:

$$\begin{array}{c}\gg \gg \underset{R\to +\infty }{\lim }{-\frac{1}{2}{e}^{-x^2}|}_0^R+\underset{R\to -\infty }{\lim }{-\frac{1}{2}{e}^{-x^2}|}_R^0\\ \\ \gg \gg -\frac{1}{2}+-\frac{1}{2}\gg \gg 0\text{(converges)}\end{array}$$

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(c)

(1) Definition of improper integral:

$${\int }_1^2\frac{1}{{(x-1)}^2}dx=\underset{R\to 1^{+}}{\lim }{\int }_R^2\frac{1}{{(x-1)}^2}dx$$

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(2) Antiderivative and limit:

$$\begin{array}{c}\gg \gg \underset{R\to 1^{+}}{\lim }{-\frac{1}{x-1}|}_R^2\\ \\ \gg \gg \underset{R\to 1^{+}}{\lim }-1--\frac{1}{R-1}\gg \gg +\infty \text{(diverges)}\end{array}$$Link to original