Name: Area Enclosed by a Curve and a Line: Illustration
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: How would you find the area enclosed between a parabola and a straight line? Watch the video to see the method as an application of definite integrals.

Question

A figure is bounded by the curve

y = |\sqrt{2} \sin \frac{π x}{4}|, y = 0, x = 2 & x = 4 .

At what angles to the positive

x - axis

straight lines must be drawn through

(4, 0)

so that these lines divide the figure into three parts of the same area.

Accepted Answer

Total area bound $A = \int_{2}^{4} \sqrt{2} \sin (\frac{π x}{4}) d x = {[- \frac{4 \sqrt{2}}{π} \cos \frac{π x}{4}]}_{2}^{4} = \frac{4 \sqrt{2}}{π}$ sq. units

Since the lines $B C & B D$ divide this area into three equal parts, so area of $∆ A B C = \frac{1}{3} \times \frac{4 \sqrt{2}}{π}$ sq. units and area of $∆ A B D = \frac{2}{3} \times \frac{4 \sqrt{2}}{π}$ sq. units.

Let slope of $B C$ be $m_{1}$ . So its equation is $(y - 0) = m_{1} (x - 4)$

Putting $x = 2$ , we get $y = - 2 m_{1}$ i.e. coordinates of $C (2, - 2 m_{1})$

So, area of $∆ A B C = \frac{1}{2} \cdot 2 \cdot (- 2 m_{1}) = - 2 m_{1}$ sq. units

$\Rightarrow - 2 m_{1} = \frac{4 \sqrt{2}}{3 π} \Rightarrow m_{1} = \frac{- 2 \sqrt{2}}{3 π}$

Let its angle with positive x-axis be $θ_{1}$ , then $m_{1} = \tan θ_{1} = \frac{- 2 \sqrt{2}}{3 π} \Rightarrow θ_{1} = π - \tan^{- 1} (\frac{2 \sqrt{2}}{3 π})$

Let slope of $B D$ be $m_{2}$ , so its equation is $(y - 0) = m_{2} (x - 4)$

Putting $x = 2$ , we get $y = - 2 m_{2}$ i.e. coordinates of $D (2, - 2 m_{2})$ .

So, area of $∆ A B D = \frac{1}{2} \cdot 2 \cdot (- 2 m_{2}) = - 2 m_{2}$ sq. units

$\Rightarrow \frac{1}{2} \cdot 2 \cdot (- 2 m_{2}) = \frac{8 \sqrt{2}}{3 π}$

$\Rightarrow m_{2} = \frac{- 4 \sqrt{2}}{3 π}$

Let its angle with positive x-axis be $θ_{2}$ , then $m_{2} = \tan θ_{2} = \frac{- 4 \sqrt{2}}{3 π} \Rightarrow θ_{2} = π - \tan^{- 1} (\frac{4 \sqrt{2}}{3 π})$

A figure is bounded by the curve $y = |\sqrt{2} \sin \frac{π x}{4}|, y = 0, x = 2 & x = 4 .$ At what angles to the positive $x - axis$ straight lines must be drawn through $(4, 0)$ so that these lines divide the figure into three parts of the same area.

Important Questions on Area under Curves

Learn and Prepare for any exam you want !

A figure is bounded by the curve y=2sinπx4, y=0, x=2 & x=4. At what angles to the positive x-axis straight lines must be drawn through 4,0 so that these lines divide the figure into three parts of the same area.

Important Questions on Area under Curves

Learn and Prepare for any exam you want !

A figure is bounded by the curve $y = |\sqrt{2} \sin \frac{π x}{4}|, y = 0, x = 2 & x = 4 .$ At what angles to the positive $x - axis$ straight lines must be drawn through $(4, 0)$ so that these lines divide the figure into three parts of the same area.