The integral Ι=∫sinx^2+2x^2cosx^2dx=xHx+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

MARKS Content Team · Accepted Answer

Correct answer: 1. Given integral Ι=∫1⋅sinx^2+x⋅2xcosx^2dx ∵∫f'g+f⋅g^(')dx=∫fg^(')dx=f⋅g+C =x⋅sinx^2+C ∴ Hx=sinx^2, whose range is -1,1 ∴ a=-1, b=1⇒a+2b=1

The integral $Ι = \int (s i n (x^{2}) + 2 x^{2} c o s (x^{2})) d x = x H (x) + C,$ (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

Solution

More Indefinite Integration Questions

The integral Ι=∫sinx2+2x2cosx2dx=xHx+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

Solution

More Indefinite Integration Questions

The integral $Ι = \int (s i n (x^{2}) + 2 x^{2} c o s (x^{2})) d x = x H (x) + C,$ (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