다중적분 전체 문제

Let f and g be two continuous functions such that $0\le {m}_{1}\le f(x)\le {M}_{1}$ for any $x\in [a,b]$ and $0\le {m}_{2}\le g(y)\le {M}_{2}$ for any $y\in [c,d].$ Show that the following inequality is true:

${m}_{1}{m}_{2}(b-a)(d-c)\le \int_{a}^{b}\,\int_{c}^{d}f(x)g(y)dy\,dx\le {M}_{1}{M}_{2}(b-a)(d-c).$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$\frac{1}{{e}^{2}}\le \iint_{R}{e}^{-{x}^{2}-{y}^{2}}dA\le 1,$ where $R=[0,1]\,\times \,[0,1]$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$\frac{{\pi }^{2}}{144}\le \iint_{R}\sin \,x\,\cos \,y\,dA\le \frac{{\pi }^{2}}{48},$ where $R=\left[\frac{\pi }{6},\frac{\pi }{3}\right]\,\times \,\left[\frac{\pi }{6},\frac{\pi }{3}\right]$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$0\le \iint_{R}{e}^{-y}\cos \,x\,dA\le {\left( \frac{\pi }{2} \right)}^{2},$ where $R=\left[0,\frac{\pi }{2}\right]\,\times \,\left[0,\frac{\pi }{2}\right]$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$0\le \iint_{R}(\ln \,x)(\ln \,y)dA\le {(e-1)}^{2},$ where $R=[1,e]\,\times \,[1,e]$

Let f and g be two continuous functions such that $0\le {m}_{1}\le f(x)\le {M}_{1}$ for any $x\in [a,b]$ and $0\le {m}_{2}\le g(y)\le {M}_{2}$ for any $y\in [c,d].$ Show that the following inequality is true:

$({m}_{1}+{m}_{2})(b-a)(d-c)\le \int_{a}^{b}\,\int_{c}^{d}[f(x)+g(y)]dy\,dx\le ({M}_{1}+{M}_{2})(b-a)(d-c).$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$\frac{2}{e}\le \iint_{R}({e}^{-{x}^{2}}+{e}^{-{y}^{2}})dA\le 2,$ where $R=[0,1]\,\times \,[0,1]$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$\frac{{\pi }^{2}}{36}\le \iint_{R}(\sin \,x+\cos \,y)dA\le \frac{{\pi }^{2}\sqrt{3}}{36},$ where $R=\left[\frac{\pi }{6},\frac{\pi }{3}\right]\,\times \,\left[\frac{\pi }{6},\frac{\pi }{3}\right]$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$\frac{{\pi }^{2}}{4}{e}^{\frac{-\pi }{2}}\le \iint_{R}(\cos \,x+{e}^{-y})dA\le \frac{{\pi }^{2}}{2},$ where $R=\left[0,\frac{\pi }{2}\right]\,\times \,\left[0,\frac{\pi }{2}\right]$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$0\le \iint_{R}(\ln x+\ln \,y)dA\le 2{(e-1)}^{2},$ where $R=[1,e]\,\times \,[1,e]$

Show that if f and g are continuous on $[a,b]$ and $[c,d],$ respectively, then

$\int_{a}^{b}\,\int_{c}^{d}[f(x)+g(y)]dy\,dx=(d-c)\int_{a}^{b}f(x)dx$

$\quad +\int_{a}^{b}\,\int_{c}^{d}g(y)dy\,dx=(b-a)\int_{c}^{d}g(y)dy+\int_{c}^{d}\,\int_{a}^{b}f(x)dx\,dy.$

In the following exercises, the function f is given in terms of double integrals.

a. Determine the explicit form of the function f. b. Find the volume of the solid under the surface $z=f(x,y)$ and above the region R. c. Find the average value of the function f on R. d. Use a computer algebra system (CAS) to plot $z=f(x,y)$ and $z={f}_{\text{ave}}$ in the same system of coordinates.

Show that $\int_{a}^{b}\,\int_{c}^{d}yf(x)+xg(y)dy\,dx=\frac{1}{2}({d}^{2}-{c}^{2})\left(\int_{a}^{b}f(x)dx\right)+\frac{1}{2}({b}^{2}-{a}^{2})\left(\int_{c}^{d}g(y)dy\right).$

Show that the average value of a function f on a rectangular region $R=[a,b]\,\times \,[c,d]$ is ${f}_{\text{ave}}\approx \frac{1}{mn}\sum_{i=1}^{m}\sum_{j=1}^{n}f\left({x}_{ij}^{*},{y}_{ij}^{*}\right),$ where $\left({x}_{ij}^{*},{y}_{ij}^{*}\right)$ are the sample points of the partition of R, where $1\le i\le m$ and $1\le j\le n.$

Use the midpoint rule with $m=n$ to show that the average value of a function f on a rectangular region $R=[a,b]\,\times \,[c,d]$ is approximated by

$${f}_{\text{ave}}\approx \frac{1}{{n}^{2}}\sum_{i,j=1}^{n}f\left(\frac{1}{2}({x}_{i-1}+{x}_{i}),\frac{1}{2}({y}_{j-1}+{y}_{j})\right).$$

Let $D$ be the region in the first quadrant bounded by $y=1-{x}^{2},y=4-{x}^{2},$ and the $x$- and $y$-axes.

a. Show that $\iint_{D}x\,dA=\int_{0}^{1}\,\int_{1-{x}^{2}}^{4-{x}^{2}}x\,dy\,dx+\int_{1}^{2}\,\int_{0}^{4-{x}^{2}}x\,dy\,dx$ by dividing the region $D$ into two regions of Type I. b. Evaluate the integral $\iint_{D}x\,dA.$

Let $D$ be the region bounded by $y=1,$ $y=x,$ $y=\ln \,x,$ and the $x$-axis.

a. Show that $\iint_{D}y\,dA=\int_{0}^{1}\,\int_{0}^{x}y\,dy\,dx+\int_{1}^{e}\,\int_{\ln \,x}^{1}y\,dy\,dx$ by dividing $D$ into two regions of Type I. b. Evaluate the integral $\iint_{D}y\,dA.$

a. Show that $\iint_{D}{y}^{2}dA=\int_{-1}^{0}\,\int_{-x}^{2-{x}^{2}}{y}^{2}dy\,dx+\int_{0}^{1}\,\int_{x}^{2-{x}^{2}}{y}^{2}dy\,dx$ by dividing the region $D$ into two regions of Type I, where $D=\{(x,y)|y\ge x,y\ge -x,y\le 2-{x}^{2}\}.$ b. Evaluate the integral $\iint_{D}{y}^{2}dA.$

Let $D$ be the region bounded by $y={x}^{2},y=x+2,$ and $y=-x.$

a. Show that $\iint_{D}x\,dA=\int_{0}^{1}\,\int_{-y}^{\sqrt{y}}x\,dx\,dy+\int_{1}^{4}\,\int_{y-2}^{\sqrt{y}}x\,dx\,dy$ by dividing the region $D$ into two regions of Type II, where $D=\{(x,y)|y\ge {x}^{2},y\ge -x,y\le x+2\}.$ b. Evaluate the integral $\iint_{D}x\,dA.$

Let $g$ be a positive, increasing, and differentiable function on the interval $[a,b].$ Show that the volume of the solid under the surface $z=g'(x)$ and above the region bounded by $y=0,$ $y=g(x),$ $x=a,$ and $x=b$ is given by $\frac{1}{2}({g}^{2}(b)-{g}^{2}(a)).$

Let $g$ be a positive, increasing, and differentiable function on the interval $[a,b],$ and let $k$ be a positive real number. Show that the volume of the solid under the surface $z=g'(x)$ and above the region bounded by $y=g(x),y=g(x)+k,x=a,$ and $x=b$ is given by $k(g(b)-g(a)).$