If $\int f'(x) e^{x^2}\, dx = (x-1)e^{x^2}+c$, then $f(x)$ is: (a) $x^3+4x^2+6x+c$ (b) $2x^3 - \frac{x^2}{2} + x + c$ (c) $\frac{2x^3}{3} - x^2 + x + c$ (d) $\frac{x^3}{2}+3x^2+4x+c$

If ∫ f'(x) ex^2 dx = (x-1)ex^2+c, then f(x) is: (a – Class 11 Mathematics

Question 113 of 129

\int f'(x) e^{x^2}\, dx = (x-1)e^{x^2}+c

(a)

x^3+4x^2+6x+c

(b)

2x^3 - \frac{x^2}{2} + x + c

(c)

\frac{2x^3}{3} - x^2 + x + c

(d)

\frac{x^3}{2}+3x^2+4x+c

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Differentiating both sides of the given equation recovers $f'(x)=2x^2-2x+1$ , and integrating that gives $f(x)=\dfrac{2x^3}{3}-x^2+x+c$ .

Since $\int f'(x)e^{x^2}\,dx = (x-1)e^{x^2}+c$ , differentiating both sides with respect to $x$ must give back the integrand:

$f'(x)e^{x^2} = \dfrac{d}{dx}\big[(x-1)e^{x^2}\big] = e^{x^2} + (x-1)(2x)e^{x^2}$ (product rule, using $\frac{d}{dx}e^{x^2}=2xe^{x^2}$ )

$= e^{x^2}\big[1+2x(x-1)\big] = e^{x^2}(2x^2-2x+1)$ .

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