I need help with integrals and I need help with the problem.
Integral of 4/sqrt(x)dx
Try differentiate
sqrt(x)
and see if what you get inspires you.
Integral of 4/sqrt(x)dx=
(x)^(-1/2)dx=(x)^(-1/2)dx
Integral of 4/sqrt(x)dx=4*Integral of
(x)^(-1/2)dx
Integral of x^n dx= x^(n+1) / (n+1)
Integral of (x)^(-1/2)dx=x^(-1/2+1) / (-1/2+1)= x^(1/2) / (1/2)= 2*sqrt(x)
Integral of 4/sqrt(x)dx=4*2*sqrt(x)+C= 8*sqroot(x)+C
i know that the derivative of that is 1/2x^(-1/2) i still don't know what to do
oh wow thank you so much!! that makes sense!!
To find the integral of 4/sqrt(x), you can use the power rule for integration. This rule states that if we have an expression of the form x^n, where n is any real number except -1, the integral can be found by adding 1 to the exponent and dividing by the new exponent.
In this case, we have 4/sqrt(x), which can be rewritten as 4x^(-1/2). Applying the power rule, we add 1 to the exponent (-1/2 + 1 = 1/2) and divide by the new exponent:
∫ 4/sqrt(x) dx = ∫ 4x^(-1/2) dx = 4 * (1/(1/2)) * x^(1/2) + C
Simplifying further:
= 8 * sqrt(x) + C
So, the integral of 4/sqrt(x) is 8 * sqrt(x) + C, where C is the constant of integration.