Question:
If $\displaystyle \int_{1}^{k} (3x^2 + 2x + 1)\,dx = 11$, then $k =$
If $\displaystyle \int_{1}^{k} (3x^2 + 2x + 1)\,dx = 11$, then $k =$
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After evaluating a definite integral, always simplify before solving for the variable.
Updated On: Mar 28, 2026
- $\dfrac{1}{2}$
- $-2$
- $-\dfrac{1}{2}$
- $2$
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The Correct Option is D
Solution and Explanation
Step 1: Integrating the given function.
\[ \int (3x^2 + 2x + 1)\,dx = x^3 + x^2 + x \]
Step 2: Applying the limits.
\[ \left[x^3 + x^2 + x\right]_{1}^{k} = 11 \] \[ (k^3 + k^2 + k) - (1 + 1 + 1) = 11 \] \[ k^3 + k^2 + k = 14 \]
Step 3: Solving for $k$.
Trying $k = 2$: \[ 2^3 + 2^2 + 2 = 8 + 4 + 2 = 14 \]
Step 4: Conclusion.
The value of $k$ is $2$.
\[ \int (3x^2 + 2x + 1)\,dx = x^3 + x^2 + x \]
Step 2: Applying the limits.
\[ \left[x^3 + x^2 + x\right]_{1}^{k} = 11 \] \[ (k^3 + k^2 + k) - (1 + 1 + 1) = 11 \] \[ k^3 + k^2 + k = 14 \]
Step 3: Solving for $k$.
Trying $k = 2$: \[ 2^3 + 2^2 + 2 = 8 + 4 + 2 = 14 \]
Step 4: Conclusion.
The value of $k$ is $2$.
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