Question:
If
\[
\int_0^1 (3x^2+2x+1)\,dx=k,
\]
then the value of \(k\) is
If
\[
\int_0^1 (3x^2+2x+1)\,dx=k,
\]
then the value of \(k\) is
Show Hint
For polynomial integrals, integrate each term separately and then substitute the upper and lower limits carefully. This avoids common arithmetic mistakes.
Updated On: Jun 10, 2026
- \(2\)
- \(3\)
- \(4\)
- \(5\)
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The Correct Option is B
Solution and Explanation
Concept:
To evaluate a definite integral, first find the antiderivative and then apply the Fundamental Theorem of Calculus:
\[
\int_a^b f(x)\,dx
=
F(b)-F(a),
\]
where \(F'(x)=f(x)\).
Step 1: Integrate the polynomial Given, \[ \int_0^1 (3x^2+2x+1)\,dx. \] Integrating term by term, \[ \int 3x^2\,dx=x^3, \] \[ \int 2x\,dx=x^2, \] \[ \int 1\,dx=x. \] Therefore, \[ \int (3x^2+2x+1)\,dx = x^3+x^2+x. \]
Step 2: Apply the limits \[ k = \left[x^3+x^2+x\right]_0^1. \] Substituting the upper limit: \[ 1^3+1^2+1 = 3. \] Substituting the lower limit: \[ 0^3+0^2+0 = 0. \] Hence, \[ k=3-0. \] \[ k=3. \] Therefore, \[ \boxed{3}. \]
Step 1: Integrate the polynomial Given, \[ \int_0^1 (3x^2+2x+1)\,dx. \] Integrating term by term, \[ \int 3x^2\,dx=x^3, \] \[ \int 2x\,dx=x^2, \] \[ \int 1\,dx=x. \] Therefore, \[ \int (3x^2+2x+1)\,dx = x^3+x^2+x. \]
Step 2: Apply the limits \[ k = \left[x^3+x^2+x\right]_0^1. \] Substituting the upper limit: \[ 1^3+1^2+1 = 3. \] Substituting the lower limit: \[ 0^3+0^2+0 = 0. \] Hence, \[ k=3-0. \] \[ k=3. \] Therefore, \[ \boxed{3}. \]
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