Question

Find the area bounded by the lines y = 1 + |x + 1|, x = - 3, x = 1, y = 0​

Answer 1

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

Consider,

$\rm \: y = 1 + |x + 1|$

We know, By definition of Modulus function,

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

So, using this,

$\begin{gathered}\begin{gathered}\bf\: |x + 1| = \begin{cases} &\sf{ \: \: x + 1, \: \: x + 1\geqslant 0} \\ \\ &\sf{ - (x + 1), \: \: x + 1 < 0} \end{cases}\end{gathered}\end{gathered}$

can be rewritten as

$\begin{gathered}\begin{gathered}\bf\: |x + 1| = \begin{cases} &\sf{ \: \: x + 1, \: \: x\geqslant - 1} \\ \\ &\sf{ - x - 1, \: \: x< - 1} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: 1 + |x + 1| = \begin{cases} &\sf{ \: \: 1 + x + 1, \: \: x\geqslant - 1} \\ \\ &\sf{ 1- x - 1, \: \: x< - 1} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: y = 1 + |x + 1| = \begin{cases} &\sf{ \: \: x + 2, \: \: x\geqslant - 1} \\ \\ &\sf{ \: \: \: \: \: - x, \: \: x< - 1} \end{cases}\end{gathered}\end{gathered}$

So, Let us first plot the line

$\rm \: y = x + 2 - - - (1)$

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

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

Now, Consider the second line

$\rm \: y = - x - - - (2)$

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 2 & \sf 2 \\ \\ \sf - 3 & \sf 3 \\ \\ \sf - 4 & \sf 4 \end{array}} \\ \end{gathered}$

[ See the attachment ]

Now, Required area is given by

$\rm \: = \: \displaystyle\int_{ - 3}^{ - 1}\rm y_{line \: 2} \: dx \: + \: \displaystyle\int_{ - 1}^{1}\rm y_{line \: 1} \: dx$

$\rm \: = \: \displaystyle\int_{ - 3}^{ - 1}\rm ( - x) \: dx \: + \: \displaystyle\int_{ - 1}^{1}\rm (x + 2) \: dx$

$\rm \: = \: \bigg[ - \dfrac{ {x}^{2} }{2} \bigg]_{ - 3}^{ - 1} + \bigg[\dfrac{ {x}^{2} }{2} + 2x\bigg]_{ - 1}^{1}$

$\rm \: = \: \dfrac{1}{2}( - 1 + 9) +\bigg[\dfrac{1}{2} + 2 - \frac{1}{2} + 2 \bigg]$

$\rm \: = \: \dfrac{1}{2}(8) + 4$

$\rm \: = \: 4 + 4$

$\rm \: = \: 8 \: square \: units$

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

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

Answer:

Concept:

Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.

The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral , for curves which are entirely on the same side of the x-axis in the given range.

If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.

Definite integral: If ∫ f(x) dx = g(x) + C, then

.

Calculation:

Let's say that the two curves are f(x, y) = x + y - 3 = 0 and g(x, y) = x2 - y - 9 = 0.

The points of their intersection are the points where f(x, y) = g(x, y).

⇒ x + y - 3 = x2 - y - 9 = 0

⇒ -x - y + 3 = x2 - y - 9 = 0

⇒ x2 + x - 12 = 0

⇒ x2 + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x + 4)(x - 3) = 0

⇒ x + 4 = 0 OR x - 3 = 0

⇒ x = -4 OR x = 3.

And, y = 3 - (-4) = 7 OR y = 3 - 3 = 0.

Hence, the curves intersect at the points B(3, 0) and C(-4, 7) as shown in the diagram below:

F1 Aman 21.11.20 Pallavi D7

The points where y = x2 - 9 cuts the x-axis (y = 0) are A(-3, 0) and B(3, 0).

The required area is the shaded part ABC = Area of BDC - Area of ADC.