Evaluate the integrals in the first four exercises.
\(\displaystyle\int^3_0\!\int^{x^2 +1}_{-x^2 +1} {\it xy}\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^1_0\!\int^{1}_{\sqrt{x}}\, (x + y)^2 {\it dy}\, {\it dx} \)
\(\displaystyle\int^1_0\!\int^{e^{2x}}_{e^x} x \ln y\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^1_0\!\int^2_1\!\int^3_2\cos\, [\pi(x+y+z)]\,{\it dx}\,{\it dy}\,{\it dz}.\)
Reverse the order of integration of the integrals in the next four exercises and evaluate.
The integral in the first exercise.
The integral in the second exercise.
The integral in the third exercise.
The integral in the fourth exercise.
Evaluate the integral \(\int^1_0\int^x_0\int^y_0(y+xz)\,dz\,{\it dy}\,{\it dx}.\)
Evaluate \(\int^1_0\int^{y^2}_ye^{x/y}\,{\it dx}\,{\it dy}.\)
Evaluate \(\int^1_0\int^{\rm (arcsin\, {\it y})/{\it y}}_0 y\cos xy\,{\it dx}\,{\it dy}.\)
Change the order of integration and evaluate \[ \int^2_0 \int^1_{y/2} (x + y)^2 {\it dx}\, {\it dy} . \]
Show that evaluating \({\intop\!\!\!\intop}_D {\it dx}\, {\it dy} \), where \(D\) is a \(y\)-simple region, reproduces the formula from one-variable calculus for the area between two curves.
Change the order of integration and evaluate \[ \int^1_0 \int^1_{y^{1/2}} (x^2 + y^3 x)\, {\it dx}\, {\it dy} . \]
Let \(D\) be the region in the \(xy\) plane inside the unit circle \(x^2 + y^2 = 1\). Evaluate \({\intop\!\!\!\intop}_D f(x,y)\, {\it dx}\, {\it dy}\) in each of the following cases:
Find \({\intop\!\!\!\intop}_D y[1-\cos\, (\pi x/4)] {\it dx}\, {\it dy} \), where \(D\) is the region in Figure 15.38.
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Evaluate the integrals in the next eight exercises. Sketch and identify the type of the region (corresponding to the way the integral is written).
\(\displaystyle\int^{\pi}_0 \int^{3 \sin x}_{\sin x} x(1 + y)\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^{1}_0 \int^{x \cos\, (\pi x/2)}_{x-1} (x^2 +xy +1)\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^{1}_{-1} \int^{(2-y)^2}_{y^{2/3}} \left(\frac{3}{2}\sqrt{x} - 2y\right)\, {\it dx}\, {\it dy}\)
\(\displaystyle\int^{2}_0 \int^{3 (\sqrt{4 - x^2})/2}_{-3 (\sqrt{4 - x^2})/2} \left({\displaystyle \frac{5}{\sqrt{2+x}}} + y^3\right) \! {\it dy}\, {\it dx} \)
\(\displaystyle\int^{1}_{0} \int^{x^2}_0 (x^2 + xy -y^2)\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^{4}_{2} \int^{y^3}_{y^2-1} 3 {\it dx}\, {\it dy} \)
\(\displaystyle\int^{1}_{0} \int^{x}_{x^2} (x + y)^2 {\it dy}\, {\it dx} \)
\(\displaystyle\int^{1}_{0} \int^{3y}_{0} e^{x+y} {\it dx}\, {\it dy} \)
In the next three exercises, integrate the given function f over the given region D.
\(f(x,y) = x-y\); \(D\) is the triangle with vertices (0, 0), (1, 0), and (2, 1).
\(f(x,y) = x^3 y + \cos x\); \(D\) is the triangle defined by \(0 \leq x \leq \pi/2, 0 \leq y \leq x\).
\(f(x,y) = x^2 + 2xy^2 +2\); \(D\) is the region bounded by the graph of \(y = -x^2 + x\), the \(x\) axis, and the lines \(x = 0\) and \(x = 2\).
In the next two exercises, sketch the region of integration, interchange the order, and evaluate.
\(\displaystyle\int^{4}_{1} \int^{\sqrt{x}}_1 (x^2 + y^2)\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^{1}_{0} \int^{1}_{1-y} (x+ y^2)\, {\it dx}\, {\it dy} \)
Show that \[ 4e^5 \leq \intop\!\!\!\intop\nolimits_{[1,3]\times [2,4]} e^{x^2 + y^2} {\it dA} \leq 4e^{25}. \]
Show that \[ 4\pi \leq \intop\!\!\!\intop\nolimits_{D} ({x^2 + y^2 +1})\, {\it dx}\, {\it dy} \leq 20 \pi, \] where \(D\) is the disk of radius 2 centered at the origin.
Suppose \(W\) is a path-connected region; that is, given any two points of \(W\) there is a continuous path joining them. If \(f\) is a continuous function on \(W\), use the intermediate-value theorem to show that there is at least one point in \(W\) at which the value of \(f\) is equal to the average of \(f\) over \(W\); that is, the integral of \(f\) over \(W\) divided by the volume of \(W\). (Compare this with the mean-value theorem for double integrals.) What happens if \(W\) is not connected?
Prove: \(\int^x_0 [ \int^t_0 F(u)\, {\it {\,d} u}] {\it {\,d} t} = \int^x_0 (x-u) F(u) \,{\it du}\).
Evaluate the integrals in the next three exercises.
\(\displaystyle\int^1_0\int^z_0\int^y_0 xy^2z^3 {\it dx}\,{\it dy}\,{\it dz}\)
\(\displaystyle\int^1_0\int^y_0\int_0^{x/\sqrt{3}}\displaystyle \frac{x}{x^2+z^2}{\it dz}\,{\it dx}\,{\it dy}\)
\(\displaystyle\int^2_1\int^z_1\int^2_{1/y}yz^2 {\it dx}\,{\it dy}\,{\it dz}\)
Write the iterated integral \(\int^1_0\int^1_{1-x}\int^1_xf(x,y,z)\,{\it dz}\,{\it dy}\,{\it dx}\) as an integral over a region in \({\mathbb R}^3\) and then rewrite it in five other possible orders of integration.
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1Readers not already familiar with this idea should review the appropriate sections of their introductory calculus text.
2Such \({\bf c}_{\it jk}\) exist by virtue of the continuity of \(f\) on \(R\); see Theorem 7 in Section 14.3.
3This states that if \(g(x)\) is continuous on \([a,b]\), then \(\int_a^b g (x)\, {\it dx} =g(c)(b-a)\) for some point \(c\in [a,b]\). The more general second mean-value theorem was proved in Section 14.2.