Integral I = e^3 x - 3x^4e^2 x + x^2 dx (where x>0) is equal to Option 1: 1/2x - 3/2x^2 + C, where C is an arbitrary constant Option 2: 1/4x - 1/2x^2 + C, where C is an arbitrary constant Option 3: 1/2x^2 - 3/4x^3 + C, where C is an arbitrary constant Option 4: 1/4x^2 - 1/2x^3 + C, where C is an arbitrary constant

The answer is '1/4x^2 - 1/2x^3 + C, where C is an arbitrary constant'. View full solution:

Integral I = 0) is equal to

14x^2 - 12x^3 + C, where C is an arbitrary constant

12x - 32x^2 + C, where C is an arbitrary constant

14x - 12x^2 + C, where C is an arbitrary constant

12x^2 - 34x^3 + C, where C is an arbitrary constant

14x^2 - 12x^3 + C, where C is an arbitrary constant

CUET Mathematics 2026 25th May Shift 1

Calculus

Medium

applied

$\dfrac{1}{2}x - \dfrac{3}{2}x^2 + C$ , where $C$ is an arbitrary constant

$\dfrac{1}{4}x - \dfrac{1}{2}x^2 + C$ , where $C$ is an arbitrary constant

$\dfrac{1}{2}x^2 - \dfrac{3}{4}x^3 + C$ , where $C$ is an arbitrary constant

$\dfrac{1}{4}x^2 - \dfrac{1}{2}x^3 + C$ , where $C$ is an arbitrary constant

✅ Correct Option: 4