The integral I=∫( sin (x2)+2 x2 cos (x2)) d x=x H(x)+C, (where C is the constant of integration). If the range of H(.x.) is [a , b] , then the value of a+2b is equal to

Question

The integral I=∫( sin (x2)+2 x2 cos (x2)) d x=x H(x)+C, (where C is the constant of integration). If the range of H(.x.) is [a , b] , then the value of a+2b is equal to (A)  (B)  (C)  (D)

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Given integral I=∫(1 ⋅ sin (x2)+x ⋅ 2 x cos (x2)) d x (∵ ∫(f prime g+f ⋅ g prime) d x=∫(f g) prime d x=f ⋅ g+C) =x ⋅ sin (x2)+C ∴ H(x)= sin (x2), whose range is [-1,1] ∴ a=-1, b=1 ⇒ a+2 b=1

Q. The integral $I = \int (sin (x^{2}) + 2 x^{2} cos (x^{2})) d x = x H (x) + C,$ (where $C$ is the constant of integration). If the range of $H (x)$ is $[a, b]$ , then the value of $a + 2 b$ is equal to

Given integral $I = \int (1 \cdot sin (x^{2}) + x \cdot 2 x cos (x^{2})) d x$
$(∵ \int (f^{'} g + f \cdot g^{'}) d x = \int (f g)^{'} d x = f \cdot g + C)$
$= x \cdot sin (x^{2}) + C$
$∴ H (x) = sin (x^{2}),$ whose range is [-1,1]
$∴ a = - 1, b = 1 \Rightarrow a + 2 b = 1$

Q. The integral I=∫(sin(x2)+2x2cos(x2))dx=xH(x)+C, (where C is the constant of integration). If the range of H(x) is [a,b] , then the value of a+2b is equal to

Given integral I=∫(1⋅sin(x2)+x⋅2xcos(x2))dx (∵∫(f′g+f⋅g′)dx=∫(fg)′dx=f⋅g+C) =x⋅sin(x2)+C ∴H(x)=sin(x2), whose range is [-1,1] ∴a=−1,b=1⇒a+2b=1

Q. The integral $I = \int (sin (x^{2}) + 2 x^{2} cos (x^{2})) d x = x H (x) + C,$ (where $C$ is the constant of integration). If the range of $H (x)$ is $[a, b]$ , then the value of $a + 2 b$ is equal to

Given integral $I = \int (1 \cdot sin (x^{2}) + x \cdot 2 x cos (x^{2})) d x$
$(∵ \int (f^{'} g + f \cdot g^{'}) d x = \int (f g)^{'} d x = f \cdot g + C)$
$= x \cdot sin (x^{2}) + C$
$∴ H (x) = sin (x^{2}),$ whose range is [-1,1]
$∴ a = - 1, b = 1 \Rightarrow a + 2 b = 1$