Find the area bounded by the curve y = 2 x 2 + 1 and the line y = 1 .

Given curve y = 2 x 2 + 1 , When y = 1 , ⇒ x = ± 1 Area = A = ∫ - 1 1 2 1 + x 2 d x A = 2 tan - 1 x - 1 1 A = 2 π 4 + π 4 = π sq.units

AOB is the positive quadrant of the ellipse x 2 a 2 + y 2 b 2 = 1 where OA = a , OB = b . Then find the area between the arc AB and the chord AB of the ellipse.

First draw figure, Now, Area of shaded part = 1 4 × area of ellipse - area of triangle △ A O B A = 1 4 πab - 1 2 ab = 1 4 ab π - 2

Sketch the graph of y = x + 3 and evaluate ∫ - 6 0 x + 3 d x

Given: The equation is y = x + 3 Some corresponding values of x and y are given in the table. x - 6 - 5 - 4 - 3 - 2 - 1 0 y 3 2 1 0 1 2 3 After plotting these points, we obtain the graph of y = x + 3 as follows. As we know that x + 3 ≤ 0 for - 6 ≤ x ≤ - 3 and x + 3 ≥ 0 for - 3 ≤ x ≤ 0 ∴ ∫ - 6 0 x + 3 d x = - ∫ - 6 - 3 x + 3 d x + ∫ - 3 0 x + 3 d x = - x 2 2 + 3 x - 6 - 3 + x 2 2 + 3 x - 3 0 ∴ ∫ - 6 0 | ( x + 3 ) | d x = - x 2 2 + 3 x - 6 - 3 + x 2 2 + 3 x - 3 0 = - - 3 2 2 + 3 - 3 - - 6 2 2 + 3 - 6 + 0 - - 3 2 2 + 3 - 3 = - - 9 2 - - 9 2 = 9 Hence, the required area is 9 square units.

The area bounded by the y - axis; y = cos x , y = sin x when 0 ≤ x ≤ π 2 is given by

First draw graph, So, Area = A = ∫ 0 π 4 cos x - sin x d x = sin x + cos x 0 π 4 = 1 2 + 1 2 - 1 = 2 - 1 Sq. units

Sketch the curves and identify the region bounded by x = 1 2 , x = 2 , y = log e x and y = 2 x . Find the area of the region.

The given curves are x = 1 2 . . . ( i ) x = 2 . . . ( ii ) y = log e x . . . ( iii ) y = 2 x . . . ( iv ) Required area = Area of A B C D A = ∫ 1 / 2 2 2 x - log x d x = 2 x log 2 - ( x log x - x ) 1 / 2 2 = 4 log 2 - 2 log 2 + 2 - 2 log 2 - 1 2 log 1 2 + 1 2 = 4 - 2 log 2 - 5 2 log 2 + 3 2 sq . units .