Area by Integration - Engineering Mathematics, Lecture notes of Engineering Mathematics

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APPLICATION

AREA by INTEGRATION

Geometric Interpretation of Area by Integration. Consider y = f(x) The region will be divided into n rectangles with equal width. Such that, Let, Si = area of the i th rectangle A = summation of the area of the n-rectangles

Suggested Steps to Determine the Area of a Plane Figure by Integration:

1. Determine the intersection points of the given boundaries or equations.
2. Graph the given functions.
3. Shade the area to be determined.
4. Consider a thin rectangle anywhere within the region, horizontal or vertical element, to represent the entire region.
5. Determine the dimensions of the rectangular element and limits of integration. Apply the integral using the extreme points as the limit of integration to determine the total area. element length width coordinates of point of intersection limits of integration horizontal xright - xleft dy ordinates read from bottom to top vertical yupper - ylower dx abscissa read from left to right
6. Set up the area of the element and evaluate the integral throughout the region.

Area by Integration Find the area of the region bounded by the line and the coordinate axes. O x y (0,4) (2,0) (x,y) y = 4 - 2x L = y w = dx

Using vertical stripping

𝐴 = 0 2 ( 4 − 2 𝑥 ) 𝑑𝑥

𝐴 = (^4)  0 2 𝑑𝑥 − (^2)  0 2 𝑥 𝑑𝑥 𝐴 = [

2 ] 0 2 𝐴 = 4 ( 2 ) − 2 2 − 0 𝐴 = 4 𝑠𝑞𝑟. 𝑢𝑛𝑖𝑡𝑠 𝐴 = 𝑎 𝑏 𝑦 𝑑𝑥 but

O x y (0,4) (2,0) y = 4 - 2x h b

CHECK: The figure formed is a right triangle

Where A^ bh 2 1   A ( 2 )( 4 ) 4 squnits 2 1   

Determine the area of the region bounded by the curve , the lines , and. O x y y = 2 y^2 = 4x (x,y) x = 0 w = dy L = x (1,2) 𝐴 = 𝐿𝑊 𝐴 = 0 2 𝑦 2 4 𝑑𝑦 𝐴 = 1 4  0 2 𝑦 2 𝑑𝑦 𝐴 =

3 3 ] 0 2 𝐴 = 1 12

3 − 0 ) 𝐴 = 2 3 𝑠𝑞𝑟. 𝑢𝑛𝑖𝑡𝑠 Point of intersection of y = 2 and the curve: If y = 2: 𝑥 = 𝑦 2 4 𝑥 = 2 2 4 𝑥 = 1 ∴ ( 1 , 2 )

Find the area of the region bounded by the curve and O V(0,0) y = x y = -x^2 (-1,-1) x y (x,yL) (x,yC) L = yC - yL w = dx

𝐴 = − 1 0

2

− 𝑥 )^ 𝑑𝑥

𝐴 = − 𝑥 3 3 − 𝑥 2 2

− 1 0 𝐴 = 0 −

− − 1 3 3 − − 1 2

𝐴 = 1 6 𝑠𝑞𝑟 .𝑢𝑛𝑖𝑡𝑠