Integral Calculus II Exercise 3.1

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Important Formulas in Integral Calculus II

Text Book Solutions for Integral Calculus I Exercise 3.2

Text Book Solutions for Integral Calculus I Exercise 3.3

Text Book Solutions for Integral Calculus I Exercise 3.4

Exercise 3.1

1. Using Integration, find the area of the region bounded the line 2y + x = 8, the x axis and the lines x = 2, x = 4.

Answer:

2x+y=8

x	0	8
y	4	0

Since the region lies above the x axis

$A= \displaystyle \int_{a}^{b} y \: dx$

$A= \displaystyle \int_{2}^{4} \dfrac{8-x}{2} \: dx\\$
$A= \displaystyle \int_{2}^{4} \dfrac{8-x}{2} \: dx\\$
$=\dfrac{1}{2}\bigg [ 8x-\dfrac{x^{2}}{2}\bigg]_{2}^{4}$
$=\dfrac{1}{2}\bigg [ \left ( 32-8 \right )-\left ( 16-2 \right )\bigg]\\$
$=\dfrac{1}{2} \left ( 24-14 \right )\\$
$=\dfrac{1}{2} \left ( 10 \right )\\$
$A= 5 \: sq\: units$

2. Find the area bounded by the lines y − 2x − 4 = 0, y = 1, y = 3 and the y-axis.

Answer:

y − 2x − 4 = 0; y= 2x+4

x	0	-2
y	4	0

Since the region lies to the left of the y axis

$A= \displaystyle \int_{c}^{d} - x \: dy$

$y-2x - 4 = 0 \\$
$-2x=4-y\\$
$x= \dfrac{y-4}{2}$

$A= \displaystyle \int_{1}^{3} -x \: dy\\$
$A= \displaystyle \int_{3}^{1} x \: dy\\$
$A= \displaystyle \int_{3}^{1} \dfrac{y-4}{2} \: dy\\$
$=\dfrac{1}{2}\displaystyle \int_{3}^{1}(y-4)\: dy \\$
$=\dfrac{1}{2} \bigg[ \dfrac{y^{2}}{2}-4y \bigg]_{3}^{1}\\$
$=\dfrac{1}{2} \bigg[\left( \dfrac{1}{2} -4\right)-\left( \dfrac{9}{2} -12\right)\bigg]\\$
$=\dfrac{1}{2} \left[ \left( -\dfrac{7}{2} \right) - \left( -\dfrac{15}{2} \right) \right]\\$
$= \dfrac{1}{2} \left( -\dfrac{7}{2} +\dfrac{15}{2} \right)\\$
$= \dfrac{1}{2} \left( \dfrac{8}{2} \right)\\$
$= \dfrac{1}{2} (4)\\$
$A = 2 \: sq. \: units$

$3. \: Calculate\; the \; area\; bounded \;by \;the \; parabola \; y^{2}=4ax\; and \; its \;latus \;rectum.$

Answer:

$y^{2}=4ax\\$
$y=2\sqrt{a}\:\sqrt{x}\\$
$A = 2\displaystyle \int_{a}^{b} y\: dx\\$
$= 2\displaystyle \int_{0}^{a} 2 \sqrt{a}\:\sqrt{x} \: dx\\$
$= 4\: \sqrt{a}\ \displaystyle \int_{0}^{a} \sqrt{x} \: dx\\$
$= 4\: \sqrt{a}\ \displaystyle \int_{0}^{a} {x^{\frac{1}{2}}} \: dx\\$
$= 4\: \sqrt{a}\ \left[ \dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{0}^{a}\\$
$= 4\: \sqrt{a}\ \left[ \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{a}\\$
$= 4\: \sqrt{a} \times \dfrac{2}{3} \left[ {x^{\frac{3}{2}}} \right]_{0}^{a}\\$
$= \dfrac{8\:\sqrt{a}}{3} \left[ {x^{\frac{3}{2}}} \right]_{0}^{a}\\$
$= \dfrac{8\:\sqrt{a}}{3} (a^{\frac{3}{2}}-0)\\$
$= \dfrac{8\:\sqrt{a}}{3} a\sqrt{a}\\$
$A = \dfrac{8\:{a^{2}}}{3} \: sq.\: units$

4. Find the area bounded by the line y = x, the x-axis and the ordinates x = 1, x = 2.

Answer:

Since the region lies above the x axis,

$A=\displaystyle \int_{a}^{b} y \: dx\\$
$=\displaystyle \int_{1}^{2} x \: dx\\$
$=\left[ \dfrac{x^{2}}{2} \right]_{1}^{2}\\$
$= 2-\dfrac{1}{2}\\$
$A = \dfrac{3}{2}\; sq. \; units$

5. Using integration, find the area of the region bounded by the line y −1 = x ,the x axis and the ordinates x = –2, x = 3.

Answer:

$y-1=x$

x	0	-1
y	1	0

$A=A_{1}+A_{2}\\$
$A_{1}= \displaystyle \int_{a}^{b}(-y)\: dx\; (Since\; region\; lies\; below\; y\; axis)\\$
$=\displaystyle \int_{-2}^{-1}(-y)\: dx\\$
$=\displaystyle \int_{-1}^{-2}(y)\: dx\\$
$=\displaystyle \int_{-1}^{-2}(x+1)\: dx\\$
$=\left[ \dfrac{x^{2}}{2}+x \right]_{-1}^{-2}\\$
$= (2-2)-(\frac{1}{2}-1)\\$
$=0+\dfrac{1}{2}\\$
$A_{1}=\dfrac{1}{2}\\$
$A_{2}= \displaystyle \int_{a}^{b}(y)\: dx\; (Since\; region\; lies\; above\; x\; axis)\\$
$=\displaystyle \int_{-1}^{3}(y)\: dx\\$
$=\displaystyle \int_{-1}^{3}(x+1)\: dx\\$
$=\left[ \dfrac{x^{2}}{2}+x \right]_{-1}^{3}\\$
$= (\dfrac{9}{2}+3)-(\frac{1}{2}-1)\\$
$=\dfrac{15}{2}+\dfrac{1}{2}\\$
$=\dfrac{16}{2}\\$
$A_{2}=8\\$
$A=A_{1}+A_{2}\\$
$=\dfrac{1}{2}+8\\$
$A=\dfrac{17}{2}\; sq. \; units$

6. Find the area of the region lying in the first quadrant bounded by the region y=4x² ; x=0; y=0; and y=4.

Answer:

$y=4x^{2}\\$
$x^{2}=\dfrac{y}{4}\\$
$x=\dfrac{\sqrt{y}}{2}\\$
$Since \: the \: region \: lies \: to \: the \: right \: of \: the \: y axis\\$
$A=\displaystyle \int_{c}^{d} x\; dy\\$
$=\displaystyle \int_{0}^{4} \dfrac{\sqrt{y}}{2}\; dy\\$
$=\dfrac{1}2{}\displaystyle \int_{0}^{4}\sqrt{y}\: dy\\$
$=\dfrac{1}2{}\displaystyle \int_{0}^{4} {y^{\frac{1}{2}}}\: dy\\$
$=\dfrac{1}{2}\left[ \dfrac{ y^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{4}\\$
$= \dfrac{1}{2}\times \dfrac{2}{3} \left[ y^{\frac{3}{2}} \right]_{0}^{4}\\$
$= \dfrac{1}{3} \left( 4^{\frac{3}{2}}-0 \right)\\$
$= \dfrac{1}{3} \times 8\\$
$A =\dfrac{8}{3}\; sq. units\\$

$7.\; Find\; the \;area \;bounded \;by \;the \;curve \;y = x^{2} \;and \;the \;line \;y = 4.$

Answer:

$y=x^{2}\\$
$x^{2}=y\\$
$x=\sqrt{y}\\$
$A=2\displaystyle \int_{c}^{d}x dy\\$
$=2\displaystyle \int_{0}^{4}\sqrt{y} \;dy\\$
$=2\displaystyle \int_{0}^{4} y^{\frac{1}{2}} \;dy\\$
$=2 \left[ \dfrac{y^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{4}\\$
$= 2\times \dfrac{2}{3} \left[ {y^{\frac{3}{2}}}\right]_{0}^{4}\\$
$= \dfrac{4}{3} (8-0)\\$
$A= \dfrac{32}{3} sq.\; units$