Question

If \(\int f^'(x)\cdot e^{x^2}\,dx = (x-1)\cdot e^{x^2}+k\), where \(k\) is constant of integration, then \(f(x) = \ldots\)

• A. \(2x^3-\frac{x^2}{2}+x+c\), where \(c\) is constant of integration.
• B. \(\frac{x^3}{2}+3x^2+4x+c\), where \(c\) is constant of integration.
• C. \(x^3+4x^2+6x+c\), where \(c\) is constant of integration.
• D. \(\frac{2x^3}{3}-x^2+x+c\), where \(c\) is constant of integration.

Hint:

Differentiate both sides to get f'(x)e^(x^2), then integrate f'(x).

Answer 1

The Correct Option is D

Step 1: Verify by differentiating each option:
Since $f'(x)$ must equal $2x^2 - 2x + 1$ (found from the derivative of $(x-1)e^{x^2}$ divided by $e^{x^2}$), differentiate the options.
(A) $f' = 6x^2 - x + 1$. No.
(B) $f' = \frac{3x^2}{2} + 6x + 4$. No.
(C) $f' = 3x^2 + 8x + 6$. No.
(D) $f' = 2x^2 - 2x + 1$. Yes.

Step 2: Derivation:
$\frac{d}{dx}[(x-1)e^{x^2}] = e^{x^2}(1 + 2x(x-1)) = e^{x^2}(2x^2-2x+1)$.

Final Answer:
Option (D). \[ \boxed{\frac{2x^3}{3}-x^2+x+c \text{ (D)}} \]