If \( \int_0^3\displaystyle (8x^3+5x^2-2x-k)\,dx=3 \), where \( k \) is constant, then \( k \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 65
Integrating term by term:
\( \displaystyle \int_0^3(8x^3+5x^2-2x-k)\,dx=\left[2x^4+\frac{5x^3}{3}-x^2-kx\right]_0^3 \)
\( \displaystyle =162+45-9-3k \)
\( \displaystyle =198-3k \)
Setting this equal to 3:
\( \displaystyle 198-3k=3 \)
\( \displaystyle 3k=195 \)
\( \displaystyle k=65 \)