Question

If \(\int _0^a(x^2-4x+1)dx = 6\), then the real value of \(a\) is

• A. \(4\)
• B. \(6\)
• C. \(3\)
• D. \(-3\)

Hint:

Integrate, set equal to 6, and factor the cubic.

Answer 1

The Correct Option is B

Step 1: Plan:
Test the options in the integral formula.

Step 2: Test:
The value of the integral is $F(a) = \frac{a^3}{3}-2a^2+a$. $F(4) = \frac{64}{3}-32+4 = -6.67$. $F(3) = 9-18+3 = -6$. $F(-3) = -9-18-3 = -30$. $F(6) = 72-72+6 = 6$.
Only $a=6$ gives the value $6$.

Final Answer:
The real value is $a = 6$, option (B). \[ \boxed{6} \]