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If ∫−1bx3 dx=154\displaystyle\int_{-1}^{b} x^3\,dx = \dfrac{15}{4}, then bb can take the value

Solution

✅ Correct Option: 3

∫−1bx3dx=b44−14=154\int_{-1}^{b}x^3dx = \frac{b^4}{4}-\frac{1}{4} = \frac{15}{4} gives b4=16b^4 = 16, so b=±2b = \pm2. Among the options, b=2b = 2.

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