# if ∫(3x^2+2x+1)dx=14, then α - Mathematics and Statistics

If int_0^alpha(3x^2+2x+1)dx=14 then alpha=

(A) 1

(B) 2

(C) –1

(D) –2

#### Solution

(B) 2

int_0^alpha (3x^2+2x+1)dx=14

[x^3+x^2+x]_0^alpha=14

alpha^3+alpha^2+alpha-14=0

(alpha-2)(alpha^2+3alpha+7)=0

But alpha^2+3alpha+7=0 does not have real roots

alpha=2

Is there an error in this question or solution?