Question

If \[ \int_{0}^{1}(3x^2+2x+1)\,dx=k, \] then the value of \(k\) is:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

For polynomial functions, integrate each term separately and then apply the limits carefully.

Answer 1

The Correct Option is B

Concept: A definite integral gives the exact area under a curve between the specified limits. For evaluating a definite integral, we first find the antiderivative and then apply the limits using: \[ \int_a^b f(x)\,dx = F(b)-F(a) \] where \(F(x)\) is an antiderivative of \(f(x)\).

Step 1: Find the antiderivative Given, \[ k= \int_0^1(3x^2+2x+1)\,dx \] Integrating term by term, \[ \int (3x^2+2x+1)\,dx = x^3+x^2+x \] Therefore, \[ k= \Big[x^3+x^2+x\Big]_0^1 \]

Step 2: Substitute the upper limit At \(x=1\), \[ 1^3+1^2+1 = 1+1+1 = 3 \]

Step 3: Substitute the lower limit At \(x=0\), \[ 0^3+0^2+0 = 0 \]

Step 4: Apply Fundamental Theorem of Calculus \[ k=3-0 \] \[ k=3 \] Final Answer: \[ \boxed{3} \]