Question:medium
$\displaystyle \int_{-6}^{0} \left[t^3 + 9t^2 + 27t + 29 + (t+3)\cos(t+3)\right] dt$ is equal to:
$\displaystyle \int_{-6}^{0} \left[t^3 + 9t^2 + 27t + 29 + (t+3)\cos(t+3)\right] dt$ is equal to:
Show Hint
Use symmetry: odd functions integrate to zero over symmetric limits.
Updated On: Apr 24, 2026
- $6$
- $12$
- $18$
- $4$
- $24$
Show Solution
The Correct Option is B
Solution and Explanation
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