find an antiderivative of the function
((t-1)^2)/sqrt(t)
that goes through the point (1,2)
i find derivative of top and bottom first and get
2t-2/(1/2*sqrt(t)) but this doesnt give right solution
then i find diervative of that again and i get 2/(-1/4 * t^(-3/2))
what am i doing wrong please
To find the antiderivative of the given function, you need to use the technique of integration. Let's break down the steps to find the antiderivative correctly.
1. Start with the given function: f(t) = ((t-1)^2)/sqrt(t).
2. Simplify the function if needed. In this case, you can expand the numerator:
f(t) = (t^2 - 2t + 1) / sqrt(t).
3. To find the antiderivative, integrate each term separately. Recall the power rule and the rule for integrating the reciprocal of a square root:
∫ t^2 / sqrt(t) dt = (2/3) * t^(3/2) + C1,
∫ -2t / sqrt(t) dt = -4/3 * t^(3/2) + C2,
∫ 1 / sqrt(t) dt = 2 * sqrt(t) + C3.
Where C1, C2, and C3 are constants representing the constant of integration.
4. Combine the individual antiderivatives to find the antiderivative of the original function:
F(t) = (2/3) * t^(3/2) - (4/3) * t^(3/2) + 2 * sqrt(t) + C,
Where C is a constant that includes the sum of the constants of integration.
5. To determine the value of the constant C and make the antiderivative pass through the point (1,2), substitute the values of t and the function into the equation:
2 = (2/3) * 1^(3/2) - (4/3) * 1^(3/2) + 2 * sqrt(1) + C.
Simplify the equation:
2 = 2/3 - 4/3 + 2 + C.
Combine like terms and solve for C:
2 = 6/3 - 4/3 + 2 + C,
2 = 4/3 + 2 + C,
2 - 4/3 - 2 = C,
C = -4/3.
6. Finally, substitute the value of C into the equation for the antiderivative:
F(t) = (2/3) * t^(3/2) - (4/3) * t^(3/2) + 2 * sqrt(t) - 4/3.
Therefore, the antiderivative of the function ((t-1)^2)/sqrt(t) that goes through the point (1,2) is:
F(t) = (2/3) * t^(3/2) - (4/3) * t^(3/2) + 2 * sqrt(t) - 4/3.