Question

Sketch the graph of y = |x-3| and hence evaluate

$\int_1^5 |x - 3| dx$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = |x - 3|$

We know, By definition of Modulus function,

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x, \: \: when \: x < 0} \\ \\ &\sf{ \: \: x, \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition, we have

$\begin{gathered}\begin{gathered}\bf\: y = |x - 3| = \begin{cases} &\sf{ - (x - 3), \: \: when \: x - 3< 0} \\ \\ &\sf{ \: \: x - 3, \: \: when \: x - 3 \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: y = |x - 3| = \begin{cases} &\sf{ 3- x, \: \: when \: x < 3} \\ \\ &\sf{ x - 3, \: \: when \: x \geqslant 3} \end{cases}\end{gathered}\end{gathered}$

So, we have two lines

$\rm \: y = 3 - x \: \: when \: x < 3$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 3 \\ \\ \sf 1 & \sf 2 \end{array}} \\ \end{gathered}$

and

$\rm \: y = x - 3 \: \: when \: x \geqslant 3$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 1 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

[ See the attachment ]

Let assume that

$\rm \: y_1 = 3 - x$

and

$\rm \: y_2 = x - 3$

Hence,

$\rm \: \displaystyle\int_1^5\rm |x - 3| \: dx$

$\rm \: = \: \displaystyle\int_1^3\rm y_1 \: dx \: + \: \displaystyle\int_3^5\rm y_2 \: dx$

$\rm \: = \: \displaystyle\int_1^3\rm (3 - x) \: dx \: + \: \displaystyle\int_3^5\rm (x - 3) \: dx$

$\rm \: = \bigg[3x - \dfrac{ {x}^{2} }{2} \bigg]_1^3 + \bigg[\dfrac{ {x}^{2} }{2} - 3x\bigg]_3^5$

$\rm \: = \bigg(9 - \dfrac{9}{2} \bigg) - \bigg(3 - \dfrac{1}{2} \bigg) + \bigg | \bigg(\dfrac{25}{2} - 15\bigg) - \bigg(\dfrac{9}{2} - 9\bigg)\bigg |$

$\rm \: = 4.5 - 2.5 + | - 2.5 + 4.5|$

$\rm \: = 2 + 2$

$\rm \: = 4 \: sq. \: units$

Hence,

$\rm\implies \:\rm \: \displaystyle\int_1^5\rm |x - 3| \: dx \: = \: 4$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}Solution−</p><p></p><p>Given function is \\ </p><p></p><p>\begin{gathered}\rm \: y = |x - 3| \\ \end{gathered}y=∣x−3∣</p><p></p><p>We know, By definition of Modulus function, \\ </p><p></p><p>\begin{gathered}\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x, \: \: when \: x < 0} \\ \\ &\sf{ \: \: x, \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}\end{gathered}∣x∣=⎩⎪⎪⎨⎪⎪⎧−x,whenx<0x,whenx⩾0 \\ </p><p></p><p>So, using this definition, we have \\ </p><p></p><p>\begin{gathered}\begin{gathered}\begin{gathered}\bf\: y = |x - 3| = \begin{cases} &\sf{ - (x - 3), \: \: when \: x - 3 < 0} \\ \\ &\sf{ \: \: x - 3, \: \: when \: x - 3 \geqslant 0} \end{cases}\end{gathered}\end{gathered}\end{gathered}y=∣x−3∣=⎩⎪⎪⎨⎪⎪⎧−(x−3),whenx−3<0x−3,whenx−3⩾0</p><p></p><p>\begin{gathered}\begin{gathered}\begin{gathered}\bf\: y = |x - 3| = \begin{cases} &\sf{ 3- x, \: \: when \: x < 3} \\ \\ &\sf{ x - 3, \: \: when \: x \geqslant 3} \end{cases}\end{gathered}\end{gathered}\end{gathered}y=∣x−3∣=⎩⎪⎪⎨⎪⎪⎧3−x,whenx<3x−3,whenx⩾3</p><p></p><p>So, we have two lines</p><p></p><p>\begin{gathered}\rm \: y = 3 - x \: \: when \: x < 3 \\ \end{gathered}y=3−xwhenx<3</p><p></p><p>➢ Pair of points of the given equation are shown in the below table.</p><p></p><p>\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 3 \\ \\ \sf 1 & \sf 2 \end{array}} \\ \end{gathered}\end{gathered}x01y32</p><p></p><p>and</p><p></p><p>\begin{gathered}\rm \: y = x - 3 \: \: when \: x \geqslant 3 \\ \end{gathered}y=x−3whenx⩾3</p><p></p><p>➢ Pair of points of the given equation are shown in the below table.</p><p></p><p>\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 1 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}\end{gathered}x43y10</p><p></p><p>[ See the attachment ]</p><p></p><p>Let assume that</p><p></p><p>\begin{gathered}\rm \: y_1 = 3 - x \\ \end{gathered}y1=3−x</p><p></p><p>and</p><p></p><p>\begin{gathered}\rm \: y_2 = x - 3 \\ \end{gathered}y2=x−3</p><p></p><p>Hence,</p><p></p><p>\begin{gathered}\rm \: \displaystyle\int_1^5\rm |x - 3| \: dx \\ \end{gathered}∫15∣x−3∣dx</p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int_1^3\rm y_1 \: dx \: + \: \displaystyle\int_3^5\rm y_2 \: dx \\ \end{gathered}=∫13y1dx+∫35y2dx</p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int_1^3\rm (3 - x) \: dx \: + \: \displaystyle\int_3^5\rm (x - 3) \: dx \\ \end{gathered}=∫13(3−x)dx+∫35(x−3)dx</p><p></p><p>\begin{gathered}\rm \: = \bigg[3x - \dfrac{ {x}^{2} }{2} \bigg]_1^3 + \bigg[\dfrac{ {x}^{2} }{2} - 3x\bigg]_3^5 \\ \end{gathered}=[3x−2x2]13+[2x2−3x]35</p><p></p><p>\rm \: = \bigg(9 - \dfrac{9}{2} \bigg) - \bigg(3 - \dfrac{1}{2} \bigg) + \bigg | \bigg(\dfrac{25}{2} - 15\bigg) - \bigg(\dfrac{9}{2} - 9\bigg)\bigg |=(9−29)−(3−21)+∣∣∣∣∣(225−15)−(29−9)∣∣∣∣∣</p><p></p><p>\rm \: = 4.5 - 2.5 + | - 2.5 + 4.5|=4.5−2.5+∣−2.5+4.5∣</p><p></p><p>\begin{gathered}\rm \: = 2 + 2 \\ \end{gathered}=2+2</p><p></p><p>\begin{gathered}\rm \: = 4 \: sq. \: units \\ \end{gathered}=4sq.units</p><p></p><p>Hence,</p><p></p><p>\begin{gathered}\rm\implies \:\rm \: \displaystyle\int_1^5\rm |x - 3| \: dx \: = \: 4 \\ \end{gathered}⟹∫15∣x−3∣dx=4</p><p></p><p>\rule{190pt}{2pt}</p><p></p><p>Additional Information :-</p><p></p><p>\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}f(x)ksinxcosxsec2xcosec2xsecxtanxcosecxcotxtanxx1ex∫f(x)dxkx+c−cosx+csinx+ctanx+c−cotx+csecx+c−cosecx+clogsecx+clogx+cex+c</p><p></p><p>$