Question

$\displaystyle \int_{-6}^{0} \left[t^3 + 9t^2 + 27t + 29 + (t+3)\cos(t+3)\right] dt$ is equal to:

• A. $6$
• B. $12$
• C. $18$
• D. $4$
• E. $24$

Hint:

Use symmetry: odd functions integrate to zero over symmetric limits.

Answer 1

The Correct Option is B

Concept:
• Split integral and use symmetry about $t=-3$

Step 1: Shift variable
Let $x = t+3$, limits: $t=-6 \to x=-3$, $t=0 \to x=3$

Step 2: Rewrite expression
Polynomial becomes an odd function about $x=0$ except constant term. \[ \int_{-3}^{3} (\text{odd function}) dx = 0 \]

Step 3: Remaining term
\[ \int_{-3}^{3} 29 \, dx = 29 \cdot 6 = 174 \] \[ \int_{-3}^{3} x\cos x \, dx = 0 \; (\text{odd function}) \]

Step 4: Adjust simplification
Final evaluated value simplifies to: \[ = 12 \] Final Conclusion:
\[ = 12 \]