value for b for which the area under the curve y = x^2 + bx between x = 0 and x = 1 is exactly equal to 10?

Question

a. 11/3

b. 2/9

c. 58/3

d. -34/3

e. -17/12

f. none of these

Answer 1

A = integral (0 to 1) [x^2 + bx]dx
A= x^3/3 +bx^2/2 evaluated from 0 to 1
A = 1/3 +b/2 = 10
b/2 = 9 2/3
b = 19 1/3

Answer 2

integrate and evaluate from 0 to 1 to get 1/3+b/2=10 solve for b to get 58/3

Answer 3

integral(x^2 + bx, from 0 to 1) = 10
1/3x^3 + 1/2bx^2, from 0 to 1 = 10
(1/3(1)^3 + 1/2(b)(1)^2) - (1/3(0)^3 + 1/2(b)(0)^2) = 10
1/3 + 1/2b - 0 = 10
1/2b = 29/3
b = 58/3