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# Find the area of the figure bounded by the given pair of curves y = x2 - x + 3 a...

### Question

Find the area of the figure bounded by the given pair of curves y = x2 - x + 3 and y = 3

### Options

A) 17/6 units (sq)

B) 7/6 units (sq)

C) 5/6 units (sq)

D) 1/6 units (sq)

### Explanation:

Area bounded by
y = x2 - x + 3 and y = 3
x2 - x + 3 = 3
x2 - x = 0
x(x -1) = 0
x= 0 and x = 1
$$\int^{1}_{0}$$(x2 - x + 3)dx = [1/3x3 - 1/2x2 + 3x]$$^1 _0$$
(1/3(1)3 - 1/2(1)2 + 3(1) - (1/3(0)3 - 1/2(0)2 + 3(0))
= (1/3 - 1/2 + 3)- 0
= $$\frac{2-3+18}{6}$$
= 17/6

## Dicussion (1)

• Area bounded by
y = x2 - x + 3 and y = 3
x2 - x + 3 = 3
x2 - x = 0
x(x -1) = 0
x= 0 and x = 1
$$\int^{1}_{0}$$(x2 - x + 3)dx = [1/3x3 - 1/2x2 + 3x]$$^1 _0$$
(1/3(1)3 - 1/2(1)2 + 3(1) - (1/3(0)3 - 1/2(0)2 + 3(0))
= (1/3 - 1/2 + 3)- 0
= $$\frac{2-3+18}{6}$$
= 17/6