I need some anti derivatives.?

Question

I have a test tommorrow on antiderivatives, I really dont understand them but im allowed a 1/2 page of notes. I confident with everything but antiderivative so ill give you a couple of the one I dont understand and if you help me with the antiderivatives. Only answer if you are positive about these, dont just guess please

1/x^2, 1/square root of x, 1/x, 5/t^2, -7x, 5t^3, x/5x

Answer 1

1/x^2, 1/square root of x, 1/x, 5/t^2, -7x, 5t^3, x/5x

1/x^2 =x^{-2} so the antiderivative is: x^{-1}/-1 + C = -1/x +C

1/square root of x = x^{-1/2}, so the antiderivative is:
x^{1/2}/1/2 + C = 2x^{1/2} + C

1/x, the antiderivative is: ln(x) + C

-7x, the antiderivative is: -7x^2 /2 + C

5t^3, the antiderviative is: 5t^4/4 + C

x/5x = 1/5, so the antiderivative is: x/5 + C .

Answer 2

The general rule for taking antiderivatives of powers goes as follows:

Remember that taking the derivative of x^n (n some constant) is simple.
f(x) = x^n
f'(x) = n x^(n-1)

When taking the antiderivative, it's like going in reverse. So if we denote F(x) to be the antiderivative, if

f(x) = x^n
F(x) = [x^(n+1)]/(n+1)

NOTE!!!!!!! The formula will NOT work if n = -1. If n = -1, then x^n = x^(-1) = 1/x. The antiderivative of 1/x is ln|x|

In other words, you're ADDING 1 to the power, and dividing by (n+1).

In your examples, your goal is to change all of them to the appropriate form. When taking the antiderivative you can ignore constants.

Let's try the first one:
f(x) = 1/x^2

Your first goal is to change it to the proper form. Remember that you can move x^2 to the numerator, and when you do that, the exponent changes to negative.

f(x) = x^(-2)/1

And now we apply the formula we said:

F(x) = [x^(-1)]/(-1) + C
F(x) = (-1)[x^(-1)] + C
F(x) = -1/x + C

Remember to ALWAYS add a constant C when taking the antiderivative.

For the second example, 1/sqrt(x), note that sqrt(x) is equal to x^(1/2), so 1/sqrt(x) = 1/[x^(1/2)], which is equal to x^(-1/2)

Therefore, if

f(x) = x^(-1/2), then
F(x) = x^(-1/2 + 1) / (-1/2 + 1) + C
F(x) = x^(1/2) / (1/2) + C
F(x) = 2x^(1/2) + C

If f(x) = 1/x, then f(x) = x^(-1). This is the only exception, and again, the answer F(x) = ln|x| + C

I'll leave it to you to do the rest; general technique is to change the form into Ax^(n), where A and n are constants.

Answer 3

these are all very straight forward. you are looking for a function whose derivative is the given function. All of these are of the form x^n. The antiderivative of x^n is (1/(n+1)*x^(n+1)

The strategy is to re-write each of the funtions so they are in this form.

1/x^2 = x^-2 with antiderivative (1/-1)*x^-1

1/sqrt(x) = x^(-.5) with antiderivative (1/.5)*x^.5

as you work the others, remember that you can just factor out the constants and multiply at the end.

Ex: ∫-7x dz = -7∫x dx = -7 [ 1/2 * x^2]

you can do the others, right?