Find the area of the figure bounded by the given pair of curves y = x

^{2}– x + 3 and y = 3-
**A.**

17/6 units (sq) -
**B.**

7/6 units (sq) -
**C.**

5/6 units (sq) -
**D.**

1/6 units (sq)

##### Correct Answer: Option A

##### Explanation

Area bounded by

y = x^{2} – x + 3 and y = 3

x^{2} – x + 3 = 3

x^{2} – x = 0

x(x -1) = 0

x= 0 and x = 1

(int^{1}_{0})(x^{2} – x + 3)dx = [^{1}/_{3}x^{3} – ^{1}/_{2}x^{2} + 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}