Evaluate the integrals in Exercises 1 to 4.
\(\displaystyle\int^3_0\!\int^{x^2 +1}_{-x^2 +1} {\it xy}\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^1_0\!\int^{1}_{\sqrt{x}}\, (x + y)^2 {\it dy}\, {\it dx} \)
\(\displaystyle\int^1_0\!\int^{e^{2x}}_{e^x} x \ln y\, {\it dy}\, {\it dx} \)
\(\displaystyle\int^1_0\!\int^2_1\!\int^3_2\cos\, [\pi(x+y+z)]\,{\it dx}\,{\it dy}\,{\it dz}.\)
Reverse the order of integration of the integrals in Exercises 5 to 8 and evaluate.
The integral in Exercise 1
The integral in Exercise 2
The integral in Exercise 3
The integral in Exercise 4
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 5.38.
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Evaluate the integrals in Exercises 17 to 24. 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 Exercises 25 to 27, 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 Exercises 28 and 29, 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 Exercises 34 to 36.
\(\displaystyle\int^1_0\int^z_0\int^y_0 xy^2z^3 {\it dx}\,{\it dy}\,{\it dz}\)
\(\displaystyle\int^1_0\int^y_0\int_0^{x/\sqrt{3}}\displaystyle \frac{x}{x^2+z^2}{\it dz}\,{\it dx}\,{\it dy}\)
\(\displaystyle\int^2_1\int^z_1\int^2_{1/y}yz^2 {\it dx}\,{\it dy}\,{\it dz}\)
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 3.3.
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 3.2.