You have two options to proceed. First, you can stick with the function as a variable of x, that is, f(x) = 1 - x^1/2, in which case we would take the derivative which gives f'(x) = -1/(2x^1/2). Then the integral becomes S sqrt(1+1/(4x)) dx = S sqrt((4x+1)/4x) dx = S sqrt(1+4x)/2x dx. Using the u-sub u=sqrt(1+4x) and du = 2dx/sqrt(1+4x), that is, dx = 1/2*u du, it becomes S (2u / (u^2-1))*(1/2*u) du = S u^2/(u^2-1) du = S 1 - 1/(u^2-1) du. Use partial fractions on the second part to get your antiderivative. Alternatively, you can utilize the inverse function here, that is, if y = 1 - x^1/2, then x = (1-y)^2. If you follow through this approach you'll get an integrand that looks similar to the one in the video but you'll need a small u substitution, that is u = 1 - y.
Considering the function x=y^2, we took the derivative of this with respect to y (given that x is a function of y here). Then dx/dy = d(y^2)/dx = 2y by the usual power rule. That is what was meant by g'(y) = 2y.
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Thx so much Sir
You're welcome.
Greatfull...
Its equivalent with f (x ) = 1 - x^1/2.
But i can't finish it... Wolud you explan / find it( equivalent above) ? 🙏🏻🙏🏻
You have two options to proceed. First, you can stick with the function as a variable of x, that is, f(x) = 1 - x^1/2, in which case we would take the derivative which gives f'(x) = -1/(2x^1/2). Then the integral becomes S sqrt(1+1/(4x)) dx = S sqrt((4x+1)/4x) dx = S sqrt(1+4x)/2x dx. Using the u-sub u=sqrt(1+4x) and du = 2dx/sqrt(1+4x), that is, dx = 1/2*u du, it becomes S (2u / (u^2-1))*(1/2*u) du = S u^2/(u^2-1) du = S 1 - 1/(u^2-1) du. Use partial fractions on the second part to get your antiderivative.
Alternatively, you can utilize the inverse function here, that is, if y = 1 - x^1/2, then x = (1-y)^2. If you follow through this approach you'll get an integrand that looks similar to the one in the video but you'll need a small u substitution, that is u = 1 - y.
Thankyou sir i understood it very well
You're very welcome.
How did u get 2y? Please!
Considering the function x=y^2, we took the derivative of this with respect to y (given that x is a function of y here). Then dx/dy = d(y^2)/dx = 2y by the usual power rule. That is what was meant by g'(y) = 2y.