If int f'(x) cdot e^{x^2} dx = (x - 1) cdot e | vedclass

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(D) Given $\int f'(x) e^{x^2} dx = (x - 1) e^{x^2} + k$.Differentiating both sides with respect to $x$:$f'(x) e^{x^2} = \frac{d}{dx} [(x - 1) e^{x^2}]

Vedclass · Accepted Answer

$\frac{2x^3}{3} - x^2 + x + c$, where $c$ is the constant of integration.

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(D) Given $\int f'(x) e^{x^2} dx = (x - 1) e^{x^2} + k$.Differentiating both sides with respect to $x$:$f'(x) e^{x^2} = \frac{d}{dx} [(x - 1) e^{x^2}]$$f'(x) e^{x^2} = (1) e^{x^2} + (x - 1) e^{x^2} (2x)$$f'(x) e^{x^2} = e^{x^2} [1 + 2x^2 - 2x]$$f'(x) = 2x^2 - 2x + 1$Integrating $f'(x)$ with respect to $x$:$f(x) = \int (2x^2 - 2x + 1) dx$$f(x) = \frac{2x^3}{3} - x^2 + x + c$Thus, option $D$ is correct.

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

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