Find the antiderivative by hand in each case.
S stands for the integral sign
A) S x*sqrt(10 + x^2) dx
So, u= 10 + x^2
du= 2xdx
du/2= xdx
(1/2) S sqrt(u) du
(1/2)*((u^(3/2))/(3/2))
(1/2)*(2/3)*(u^(3/2))
(1/3)*(u^(3/2))
= (1/3)*(10 + x^2)^(3/2) correct/incorrect?
B) S (x/(sqrt(2 - 3x)))dx
S (x)*(2 - 3x)^(-1/2)
u= 2 - 3x
du= -3dx
du/-3=dx
x= -((u - 2)/3)
=(-1/3) S (-((u - 2)/3))*(u^(-1/2))
=(-1/3) S ((-((u^(1/2)/3)-((2u^(-1/2))/3))
=(-1/3)*[((2/9)*(u^(3/2))-((4/3)*(u^(1/2))]
=[(-2/27)*((2 - 3x)^(3/2)) + (4/3)*((2 - 3x)^(1/2))] correct/incorrect?
A) To find the antiderivative of the function `x*sqrt(10 + x^2)`, you can use the substitution method.
Let's make the substitution `u = 10 + x^2`. Taking the derivative of both sides with respect to `x`, we have `du = 2xdx` or `dx = (du)/(2x)`.
Then, substitute `u` and `dx` back into the integral:
`S x*sqrt(10 + x^2) dx`
`= S (x*sqrt(u))(du/(2x))`
`= (1/2) S sqrt(u) du`
Now, we can integrate the new expression with respect to `u`:
`= (1/2) * (u^(3/2))/(3/2)`
`= (u^(3/2))/3`
`= (10 + x^2)^(3/2) / 3`
Therefore, the antiderivative of `x*sqrt(10 + x^2)` is `(10 + x^2)^(3/2) / 3`.
B) To find the antiderivative of the function `(x/(sqrt(2 - 3x)))`, you can again use the substitution method.
Let's make the substitution `u = 2 - 3x`. Taking the derivative of both sides with respect to `x`, we have `du = -3dx` or `dx = -(du)/3`.
Then, substitute `u` and `dx` back into the integral:
`= S (x)*(2 - 3x)^(-1/2)`
`= S (-((u - 2)/3))*(u^(-1/2))`
`= (-1/3) S ((-((u^(1/2))/3)-((2u^(-1/2))/3))`
`= (-1/3) * [((2/9)*(u^(3/2))-((4/3)*(u^(1/2))]`
Now, substitute `x` back in terms of `u`:
`= (-1/3) * [((2/9)*(2 - 3x)^(3/2))-((4/3)*(2 - 3x)^(1/2))]`
Therefore, the antiderivative of `(x/(sqrt(2 - 3x)))` is `(-2/27)*((2 - 3x)^(3/2)) + (4/3)*((2 - 3x)^(1/2))`.