PROBLEM 1: Compute the area of the enclosed region bounded by the graphs of the equations y = x, y = 2x, and x = 4.

SOLUTION 1: Compute the area of the enclosed region bounded by the graphs of the equations y = x, y = 2x and x = 4 . Begin by finding the points of intersection of the two graphs. From y = x and y = 2x we get that

x = 2x ⟶ −x = 0 ⟶ −x = 0 ⟶ x = 0 ⟶

Using vertical cross-sections we get that

AREA = ∫_{0}^{4}(top − bottom) dx

= ∫_{0}^{4}(2x − x) dx

= ∫_{0}^{4} x dx

= \frac{x^{2}}{2} |_{0}^{4}

= \frac{4^{2}}{2} − \frac{0^{2}}{2}

= 8 − 0 = 8

PROBLEM 2: Compute the area of the enclosed region bounded by the graphs of the equations y = x^{2} and y = x + 2 .

SOLUTION 2: Compute the area of the enclosed region bounded by the graphs of the equations y = x^{2} and y = x + 2 . Begin by finding the points of intersection of the two graphs. From y = x^{2} and y = x + 2 we get that

x^{2} = x + 2 ⟶ x^{2} − x − 2 = 0 ⟶ (x − 2)(x + 1) = 0 ⟶ x = 2 or x = −1

Using vertical cross-sections we get that

AREA = ∫_{−1}^{2}(top − bottom) dx

= ∫_{−1}^{2}((x + 2) − x^{2}) dx

= (\frac{x^{2}}{2} + 2x − \frac{x^{3}}{3} )|_{−1}^{2}

= (\frac{2^{2}}{2} + 2(2) − \frac{2^{3}}{3} ) − (\frac{(−1)^{2}}{2} + 2(−1) − \frac{(−1)^{3}}{3} )

= (2 + 4 − \frac{8}{3}) − (\frac{1}{2} − 2 + \frac{1}{3})

= (6 − \frac{8}{3}) − (\frac{3}{6} − \frac{12}{6} + \frac{2}{6})

= (\frac{36}{6} − \frac{16}{6}) − (\frac{−7}{6})

= \frac{20}{6} + \frac{7}{6}

= \frac{27}{6} .

PROBLEM 3: Compute the area of the enclosed region bounded by the graphs of the equations y = e^{x}, y = e^{−x}, and x = ln 3 .

SOLUTION 3: Compute the area of the enclosed region bounded by the graphs of the equations y = e^{x}, y = e^{−x}, and x = ln 3 . Begin by finding the points of intersection of the two graphs. From y = e^{x} and y = e^{−x} we get that

e^{x} = e^{−x} ⟶ e^{2x} = 1 ⟶ x = 0 ⟶ Using vertical cross-sections we get that

AREA = ∫_{0}^{ln 3}(top − bottom) dx

= ∫_{0}^{ln 3}(e^{x} − e^{−x}) dx

= (e^{x} − (−e^{−x}))|_{0}^{ln 3}

= (e^{x} + e^{−x})|_{0}^{ln 3}

= (e^{ln 3} + e^{− ln 3}) − (e^{0} + e^{−0})

= (3 + \frac{1}{3}) − (1 + 1)

= (\frac{10}{3}) − (2)

= \frac{10}{3} − \frac{6}{3}

= \frac{4}{3}

PROBLEM 4: Compute the area of the enclosed region bounded by the graphs of the equations y = x^{2} and y = −x^{2} + 4x + 6 .

SOLUTION 4: Compute the area of the enclosed region bounded by the graphs of the equations y = x^{2} and y = −x^{2} + 4x + 6 . Begin by finding the points of intersection of the two graphs. From y = x^{2} and y = −x^{2} + 4x + 6 we get that

x^{2} = −x^{2} + 4x + 6 ⟶ 2x^{2} − 4x − 6 = 0 ⟶ x^{2} − 2x − 3 = 0 ⟶ (x − 3)(x + 1) = 0 ⟶ x = 3 or x = −1 Using vertical cross-sections we get that

AREA = ∫_{−1}^{3}(top − bottom) dx

= ∫_{−1}^{3}((−x^{2} + 4x + 6) − (x^{2})) dx

= ∫_{−1}^{3}(−x^{2} + 4x + 6) dx

= (\frac{−2x^{3}}{3} + \frac{4x^{2}}{2} + 6x)|_{−1}^{3}

= ( − \frac{2}{3}x^{3} + 2x^{2} + 6x)|_{−1}^{3}

= ( − \frac{2}{3}(3)^{3} + 2(3)^{2} + 6(3)) − ( − \frac{2}{3}(−1)^{3} + 2(−1)^{2} + 6(−1))

= ( − 18 + 18 + 18) − (\frac{2}{3} + 2 − 6)

= (18) − (\frac{2}{3} − \frac{6}{3} − \frac{18}{3})

= (18) − ( − \frac{22}{3})

= 18 + \frac{22}{3}

= \frac{54}{3} + \frac{22}{3}

= \frac{76}{3}

PROBLEM 5: Compute the area of the enclosed region bounded by the graphs of the equations y = x^{3} + x^{2} and y = 3x^{2} + 3x . SOLUTION 2: Compute the area of the enclosed region bounded by the graphs of the equations y = x^{3} + x^{2} and y = 3x^{2} + 3x . Begin by finding the points of intersection of the two graphs. From y = x^{3} + x^{2} and y = 3x^{2} + 3x we get that x^{3} + x^{2} = 3x^{2} + 3x ⟶ x^{3} − 2x^{2} − 3x = 0 ⟶ x(x^{2} − 2x − 3) = 0 ⟶ x(x − 3)(x + 1) = 0 ⟶ x = 0, x = 3, or x = −1

Using vertical cross-sections we get that

AREA = ∫_{−1}^{0}(top − bottom) dx + ∫_{0}^{3}(top − bottom) dx

= ∫_{−1}^{0}((x^{3} + x^{2}) − (3x^{2} + 3x)) dx + ∫_{0}^{3}((3x^{2} + 3x) − (x^{3} + x^{2})) dx

= ∫_{−1}^{0}(x^{3} − 2x^{2} − 3x) dx + ∫_{0}^{3}(−x^{3} + 2x^{2} + 3x) dx

= (\frac{x^{4}}{4} − \frac{2x^{3}}{3} − \frac{3x^{2}}{2} )|_{−1}^{0} + ( − \frac{x^{4}}{4} + \frac{2x^{3}}{3} + \frac{3x^{2}}{2} )|_{0}^{3}

= (\frac{0^{4}}{4} − \frac{2(0)^{3}}{3} − \frac{3(0)^{2}}{2} ) − (\frac{(−1)^{4}}{4} − \frac{2(−1)^{3}}{3} − \frac{3(−1)^{2}}{2} ) + ( − \frac{3^{4}}{4} + \frac{2(3)^{3}}{3} + \frac{3(3)^{2}}{2} )

−( − \frac{0^{4}}{4} + \frac{2(0)^{3}}{3} + \frac{3(0)^{2}}{2} )

= (0) − (\frac{1}{4} + \frac{2}{3} − \frac{3}{2}) + ( − \frac{81}{4} + 18 + \frac{27}{2}) − (0)

= −(\frac{3}{12} + \frac{8}{12} − \frac{18}{12}) + ( − \frac{81}{4} + \frac{72}{4} + \frac{54}{4})

= −( − \frac{7}{12}) + (\frac{45}{4})

= \frac{7}{12} + \frac{45}{4}

= \frac{7}{12} + \frac{135}{12}

= \frac{142}{12}

= \frac{71}{6}

PROBLEM 6: Compute the area of the enclosed region bounded by the graphs of the equations y = ln x, y = 1, and x = e^{2} .

SOLUTION 6: Compute the area of the enclosed region bounded by the graphs of the equations y = ln x, y = 1 and y = e^{2} . Begin by finding the points of intersection of the two graphs. From y = ln x and y = 1 we get that

ln x = 1 ⟶

e^{ln x} = e^{1}

x = e

Using vertical cross-sections we get that

AREA = ∫_{e}^{e^{2}}(top − bottom) dx

= ∫_{e}^{e^{2}}(ln x − 1) dx

= ∫_{e}^{e^{2}} ln x dx − ∫_{e}^{e^{2}} 1 dx

Let A = ∫ ln x dx.

Use integration by parts. Let u = ln x and dv = dx such that du = \frac{1}{x} dx and v = x.

A = x ln x − ∫ (\frac{1}{x})(x) dx

= x ln x − ∫ dx

= x ln x − x + C

A − ∫_{e}^{e^{2}} 1 dx = (x ln x − x − x)|_{e}^{e^{2}}

= (x ln x − 2x)|_{e}^{e^{2}}

= (e^{2} ln e^{2} − 2e^{2}) − (e ln e − 2e)

= (e^{2}(2) − 2e^{2}) − (e − 2e)

= (0) − ( − e) = e

PROBLEM 7: Compute the area of the enclosed region bounded by the graphs of the equations y = cos x, y = sin x, and x = 0 .

SOLUTION 7: Compute the area of the enclosed region bounded by the graphs of the equations y = cos x, y = sin x and x = 0 . Begin by finding the points of intersection of the two graphs. From y = x^{2} and y = x + 2 we get that

cos x = sin x ⟶ \frac{cos x}{sin x} = 1 ⟶ cot x = 1 ⟶ x = \frac{π}{4}

Using vertical cross-sections we get that

AREA = ∫_{0}^{π/4}(top − bottom) dx

= ∫_{0}^{π/4}(cos x − sin x) dx

= ( sin x − (− cos x))|_{0}^{π/4}

= ( sin x + cos x)|_{0}^{π/4}

= ( sin \frac{π}{4} + cos \frac{π}{4} ) − ( sin 0 + cos 0)

= (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) − (0 + 1)

= \sqrt{2} − 1

PROBLEM 8: Compute the area of the enclosed region bounded by the graphs of the equations y = 8/x, y = 2x, and y = 2 .

SOLUTION 8: Compute the area of the enclosed region bounded by the graphs of the equations y = \frac{8}{x}, y = 2x and y = 2 . Begin by finding the points of intersection of the two graphs. From y = \frac{8}{x} and y = 2x we get that

\frac{8}{x} = 2x ⟶ 8 = 2x^{2} ⟶ 4 = x^{2} ⟶ x = 2 or x = −2 ⟶ y = 4 or y = −4

Using horizontal cross-sections such that x = \frac{8}{y} and x = \frac{y}{2} we get that

AREA = ∫_{2}^{4}(right − left) dy

∫_{2}^{4}(\frac{8}{y} − \frac{y}{2} ) dy

= (8 ln y − \frac{y^{2}}{4} )|_{2}^{4}

= (8 ln 4 − \frac{4^{2}}{4} ) − (8 ln 2 − \frac{2^{2}}{4} )

= (8 ln 4 − 4) − (8 ln 2 − 1)

= (8 ln 2^{2} − 4) − (8 ln 2 − 1)

= (16 ln 2 − 4) − (8 ln 2 − 1) = 8 ln 2 − 3

PROBLEM 9: Compute the area of the enclosed region bounded by the graphs of the equations y = ln x and y = (ln x)^{2} .

SOLUTION 9: Compute the area of the enclosed region bounded by the graphs of the equations y = ln x and y = (ln x)^{2} . Begin by finding the points of intersection of the two graphs. From y = ln x and y = (ln x)^{2} we get that

ln x = (ln x)^{2} ⟶ ln x − (ln x)^{2} = 0 ⟶ ln x(1 − ln x) ⟶ ln x = 0 or ln x = 1 ⟶ x = 1 or x = e

Using vertical cross-sections we get that AREA = ∫_{1}^{e}(top − bottom) dx

= ∫_{1}^{e}(ln x − (ln x)^{2}) dx

= ∫_{1}^{e} ln x dx − ∫_{1}^{e}(ln x)^{2} dx

Let A = ∫_{1}^{e} ln x dx and B = ∫ (ln x)^{2} dx.

(Recall from PROBLEM 6 that ∫ ln x dx = x ln −x + C )

A = (x ln x − x)|_{1}^{e}

= (e ln e − 1 ln 1) − (e − 1)

= (e − 0) − (e − 1) = 1

Use integration by parts. Let u = (ln x)^{2} and dv = dx such that du = 2 ln x(\frac{1}{x}) dx and v = x.

B = (x(ln x)^{2})|_{1}^{e} − ∫_{1}^{e} 2 ln x dx

= (x(ln x)^{2})|_{1}^{e} − 2(x ln x − x)|_{1}^{e}

= (e(ln e)^{2} − 1(ln 1)^{2}) − 2((e ln e − e) − (1 ln 1 − 1))

= ((e − 0)) − 2((e − e) − (0 − 1)) = e − 2 A − B = 1 − (e − 2) = 3 − e

PROBLEM 10: Compute the area of the enclosed region bounded by the graphs of the equations y = tan^{2} x, y = 0 and x = 1 .

SOLUTION 10: Compute the area of the enclosed region bounded by the graphs of the equations y = tan^{2} x, y = 0, and x = 1 . Begin by finding the points of intersection of the two graphs. From y = tan^{2} x and y = 0 we get that

tan^{2} x = 0 ⟶ tan x = 0 ⟶ x = 0

Using vertical cross-sections we get that

AREA = ∫_{0}^{1}(top − bottom) dx

= ∫_{0}^{1} tan^{2} x dx

= ∫_{0}^{1}(sec^{2} x − 1) dx

= ( tan x − x)|_{0}^{1}

= ( tan 1 − 1) − ( tan 0 − 0)

= tan 1 − 1

PROBLEM 11: Compute the area of the enclosed region bounded by the graphs of the equations y = x, y = 2x, and y = 6 − x .

SOLUTION 11: Compute the area of the enclosed region bounded by the graphs of the equations y = x, y = 2x, and y = 6 − x . Begin by finding the points of intersection of the two graphs. From y = x and y = 2x we get that

x = 2x ⟶ −x = 0 ⟶ x = 0

From y = x and y = 6 − x we get that x = 6 − x ⟶ 2x = 6 ⟶ x = 3

From y = 2x and y = 6 − x we get that 2x = 6 − x ⟶ 3x = 6 ⟶ x = 2

Using vertical cross-sections we get that

AREA = ∫_{0}^{2}(top − bottom) dx + ∫_{2}^{3}(top − bottom) dx

= ∫_{0}^{2}(2x − x) dx + ∫_{2}^{3}((6 − x) − x) dx

= ∫_{0}^{2} x dx + ∫_{2}^{3}(6 − 2x) dx

= \frac{x^{2}}{2} |_{0}^{2} + (6x − \frac{2x^{2}}{2} )|_{2}^{3}

= \frac{x^{2}}{2} |_{0}^{2} + (6x − x^{2})|_{2}^{3}

= (\frac{2^{2}}{2} − \frac{0^{2}}{2} ) + ((6 · 3 − 3^{2}) − (6 · 2 − 2^{2}))

= (2 − 0) + (9 − 8)

= 2 + 1

= 3

PROBLEM 12: Compute the area of the enclosed region bounded by the graphs of the equations y = sin \sqrt{x}, y = 0, and x = π/2 .

SOLUTION 12: Compute the area of the enclosed region bounded by the graphs of the equations y = sin \sqrt{x}, y = 0, and x = π/2. Begin by finding the points of intersection of the two graphs. From y = sin \sqrt{x} and y = 0 we get that

sin \sqrt{x} = 0 ⟶

x = 0 Using vertical cross-sections we get that

AREA = ∫_{0}^{π/2}(top − bottom) dx

= ∫_{0}^{π/2}(sin \sqrt{x}) dx

Use integration by parts. Let u = sin \sqrt{x} and dv = dx such that du = cos \sqrt{x} · \frac{1}{2\sqrt{x}} dx and v = x.

= x sin \sqrt{x}|_{0}^{π/2} − ∫_{0}^{π/2} cos \sqrt{x} · \frac{1}{2\sqrt{x}} · x dx

= ((\frac{π}{2} ) sin \sqrt{\frac{π}{2}} − (0) sin \sqrt{0}) − ∫_{0}^{π/2} cos \sqrt{x} · \frac{1}{2\sqrt{x}} · x dx

= \frac{π}{2} sin \sqrt{\frac{π}{2}} − ∫_{0}^{π/2} cos \sqrt{x} · \frac{1}{2\sqrt{x}} · x dx

Use substitution. Let

w = \sqrt{x} so that

dw = \frac{1}{2\sqrt{x}} dx = \frac{\sqrt{x}}{2x} dx x dw = \frac{\sqrt{x}}{2} dx ⟶ w^{2} dw = \frac{\sqrt{x}}{2} dx

= \frac{π}{2} sin \sqrt{\frac{π}{2}} − ∫_{0}^{√\overline{π/2}} cos w · w^{2} dw

Use integration by parts. Let u′ = w^{2} and dv′ = cos w dw such that du′ = 2w dw and v′ = sin w.

= \frac{π}{2} sin \sqrt{\frac{π}{2}} − (w^{2} sin w|_{0}^{√\overline{π/2}} − ∫_{0}^{√\overline{π/2}} 2w sin w dw)

= \frac{π}{2} sin \sqrt{\frac{π}{2}} − w^{2} sin w|_{0}^{√\overline{π/2}} + ∫_{0}^{√\overline{π/2}} 2w sin w dw

= \frac{π}{2} sin \sqrt{\frac{π}{2}} − ((\sqrt{\frac{π}{2}} )^{2} sin \sqrt{\frac{π}{2}} − (0)^{2} sin 0) + ∫_{0}^{√\overline{π/2}} 2w sin w dw

= \frac{π}{2} sin \sqrt{\frac{π}{2}} − \frac{π}{2} sin \sqrt{\frac{π}{2}} + ∫_{0}^{√\overline{π/2}} 2w sin w dw

= ∫_{0}^{√\overline{π/2}} 2w sin w dw Use integration by parts. Let u” = 2w and dv” = sin w dw such that du” = 2 dw and v” = − cos w

= −2w cos w|_{0}^{√\overline{π/2}} + ∫_{0}^{√\overline{π/2}} 2 cos w dw = −2((\sqrt{\frac{π}{2}} ) cos \sqrt{\frac{π}{2}} − (0) cos 0) + 2 sin w|\sqrt{\frac{π}{2}}_{0} = −2(\sqrt{\frac{π}{2}} ) cos \sqrt{\frac{π}{2}} + 2( sin \sqrt{\frac{π}{2}} − sin 0)

= −\sqrt{2π} cos \sqrt{\frac{π}{2}} + 2 sin \sqrt{\frac{π}{2}}

PROBLEM 13: Compute the area of the enclosed region bounded by the graphs of the equations x = y^{2} and x = 4 . SOLUTION 13: Compute the area of the enclosed region bounded by the graphs of the equations x = y^{2} and x = 4 . Begin by finding the points of intersection of the two graphs. From x = y^{2} and x = 4 we get that

y^{2} = 4 ⟶

y = −2 or y = 2 Using horizontal cross-sections we get that

AREA = ∫_{−2}^{2}(right − left) dy

= ∫_{−2}^{2}(4 − y^{2}) dy

= (4y − \frac{y^{3}}{3} )|_{−2}^{2}

= (4 · 2 − \frac{2^{3}}{3} ) − (4 · (−2) − \frac{(−2)^{3}}{3} )

= (8 − \frac{8}{3}) − ( − 8 + \frac{8}{3})

= 16 − \frac{16}{3}

= \frac{48}{3} − \frac{16}{3}

= \frac{32}{3}

PROBLEM 14: Compute the area of the enclosed region bounded by the graphs of the equations x = y + 3 and x = y^{2} − y . SOLUTION 14: Compute the area of the enclosed region bounded by the graphs of the equations x = y + 3 and x = y^{2} − y . Begin by finding the points of intersection of the two graphs. From x = y + 3 and x = y^{2} − y we get that

y + 3 = y^{2} − y ⟶ 0 = y^{2} − 2y − 3 ⟶ 0 = (y − 3)(y + 1) ⟶ y = 3 or y = −1

Using horizontal cross-sections we get that

AREA = ∫_{−1}^{3}(right − left) dy

= ∫_{−1}^{3}((y + 3) − (y^{2} − y)) dy

= ∫_{−1}^{3}(−y^{2} + 2y + 3) dy

= ( − \frac{y^{3}}{3} + \frac{2y^{2}}{2} + 3y)|_{−1}^{3}

= ( − \frac{y^{3}}{3} + y^{2} + 3y)|_{−1}^{3}

= ( − \frac{3^{3}}{3} + 3^{2} + 3(3)) − ( − \frac{(−1)^{3}}{3} + (−1)^{2} + 3(−1))

= ( − 9 + 9 + 9) − (\frac{1}{3} + 1 − 3)

= ( − 9 + 9 + 9) − (\frac{1}{3} + \frac{3}{3} − \frac{9}{3})

= (9) − ( − \frac{5}{3})

= \frac{27}{3} + \frac{5}{3}

= \frac{32}{3}

PROBLEM 15: Compute the area of the enclosed region bounded by the graphs of the equations x = y^{3} and x = y^{2} + 2y.

SOLUTION 15: Compute the area of the enclosed region bounded by the graphs of the equations x = y^{3} and x = y^{2} + 2y . Begin by finding the points of intersection of the two graphs. From x = y^{3} and x = y^{2} + 2y we get that

y^{3} = y^{2} + 2y ⟶ y^{3} − y^{2} − 2y = 0 ⟶ y(y^{2} − y − 2) = 0 ⟶ y(y − 2)(y + 1) = 0 ⟶ y = 0, y = 2, or y = −1

Using horizontal cross-sections we get that

AREA = ∫_{−1}^{0}(right − left) dy + ∫_{0}^{2}(right − left) dy

= ∫_{−1}^{0}(y^{3} − (y^{2} + 2y)) dy + ∫_{0}^{2}(y^{2} + 2y − y^{3}) dy

= ∫_{−1}^{0}(y^{3} − y^{2} − 2y) dy + ∫_{0}^{2}(y^{2} + 2y − y^{3}) dy

= (\frac{y^{4}}{4} − \frac{y^{3}}{3} − \frac{2y^{2}}{2} )|_{−1}^{0} + (\frac{y^{3}}{3} + \frac{2y^{2}}{2} − \frac{y^{4}}{4} )|_{0}^{2}

= (\frac{y^{4}}{4} − \frac{y^{3}}{3} − y^{2})|_{−1}^{0} + (\frac{y^{3}}{3} + y^{2} − \frac{y^{4}}{4} )|_{0}^{2}

= (\frac{0^{4}}{4} − \frac{0^{3}}{3} − 0^{2}) − (\frac{(−1)^{4}}{4} − \frac{(−1)^{3}}{3} − (−1)^{2}) + (\frac{2^{3}}{3} + 2^{2} − \frac{2^{4}}{4} ) − (\frac{0^{3}}{3} + 0^{2} − \frac{0^{4}}{4} )

= (0) − (\frac{1}{4} + \frac{1}{3} − 1) + (\frac{8}{3} + 4 − 4) − (0)

= −( − \frac{7}{12}) + (\frac{8}{3})

= \frac{7}{12} + \frac{8}{3}

= \frac{7}{12} + \frac{32}{12}

= \frac{39}{12}

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