Find the area bounded by the curve (y = x ) between (x = 0 ) and (x = 2 ).

As per the given figure, we need to find the area of the region : rgb 0,0.39215686274509803,0 (-15.16,-11.76) rectangle (7.16,7.2); plot( , sin((( ))*180/pi) ); (0,0.3) node O ; (1.5,0) node A ; (3.14,0.3) node B ; (4.5,-1.3) node C ; (6.3,0.3) node D ; OABCD = Ar.(OAB)+Ar.(BCD) Let us find the area of OAB first : = 0 sinx dx = [-cosx] 0 = -[cos - cos0]= -[-1-1]=2 Now let us find the area of BCD = 2 sinxdx =-[cosx] 2 =-[1-(-1)]= -2 as we know that area can't be negative therefore area of BCD = 2 Hence area of OAB +BCD = 2+2=4 Hence the required area = 4 sq units

Application of Integrals — Class 12 Maths Solution

exemplar la LA NCERT Exemp. Ex. 1.3, Q. 17, Page 177

Question

Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Step-by-step Solution

As per the given figure, we need to find the area of the region :

OABCD = Ar.(OAB)+Ar.(BCD)

Let us find the area of OAB first :

= $\int^{\pi}_0 sinx dx$
$= [-cosx]^{\pi}_0$

$= -[cos\pi - cos0]= -[-1-1]=2$

Now let us find the area of BCD
=$\int^{2\pi}_{\pi}sinxdx$

$=-[cosx]^{2\pi}_{\pi}$
$=-[1-(-1)]= -2$

as we know that area can't be negative therefore area of BCD = 2

Hence area of OAB +BCD = 2+2=4

Hence the required area = 4 sq units

← All Application of Integrals solutions Practice tests →

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Integrals. Curated by Sachin Sharma. Free for all students.