antiderivatives?

Question

what is the antiderivative of 6*square root x and 5/t^2

Answer 1

So you want to calculate the antiderivative of

f(x) = 6*sqrt(x).

The first thing to do is convert this into something we can actually work with. Note that sqrt(x) = x^(1/2). Let's do that.

f(x) = 6x^(1/2)

Now, note that the antiderivative of x^n is equal to [x^(n+1)]/(n+1). This is like the opposite of the power rule for taking derivatives. Thus, what we do is that "reverse power rule". Remember, when taking the antiderivative, to LEAVE ALL CONSTANTS ALONE. In the above case, we're going to leave the 6 alone, and solve. Let's denote the antiderivative to be F(x).

F(x) = 6 [ {x^(1/2 + 1)}/{(1/2 + 1} ] + C

(Side note: You always have to add a constant C when taking the general antiderivative)

F(x) = 6 [ x^(3/2) / (3/2) ] + C
F(x) = 6 [2/3 x^(3/2)] + C
F(x) = 4 x^(3/2) + C

To solve for the antiderivative of 5/(t^2), what you have to do is bring the t^2 to the numerator, and when doing so, you change the exponent to its negative. Therefore

5/(t^2) = 5t^(-2).

I'll leave it up to you to do the rest. Just use the advice I've given above.

Answer 2

1)
antideriv. 6 * sqrt x dx

= 6 antideriv. sqrt x dx

= 6 antideriv. x^1/2 dx

= 6 (x^ 1/2 + 2/2)/3/2 + C

= 6 * 2/3 x^3/2 + C

= 4x^3/2 + C

Verification:

(4x^3/2)'

4* 3/2 x^3/2 -2/2

= 6 x^1/2

= 6*sqrt x

***********************

2)

antideriv. 5/t^2 dt

= 5 andideriv. 1/t^2 dt

= 5 antideriv. t^ -2 dt

= 5 antideriv, (t ^-2 +1)/(-2 +1) dt

= 5 t^ -1/-1 + C

= -5/t + C


Verification:

(-5/t)' = -5t^ -1 = 5t^ -2 = 5/t^2

**********************

Answer 3

^ means to the power of

Antiderivative of 6*sqrt x

6 sqrt x is the same as 6x^(1/2)

Using the power rule, the antiderivative is [6*x^(1+1/2)]/(1+1/2)+C
This equals 4x^(3/2)+C

Answer 4

V is square root:

6Vx= 6x^1/2= 6(x^1/2), you don't need to find antiderivative of 6 following the Constant theorem of Derivatives

x-1=1/2 x=3/2 3/2(y)=1 y=2/3, dy/dx x^1/2= 2/3x^3/2

therefore the antiderivative of 6Vx or 6x^1/2=

6(2/3x^3/2)+ C= 12/3(x^3/2) + C= 4(x^3/2) + C, end result
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5/t^2= 5t^-2= 5(t^-2),you don't need to find antiderivative of 5 following the Constant theorem of Derivatives

x-1=-2 x=-1 y(-1)=1 y=-1, as such dy/dx t^-2= -1t^-1

therefore the antiderivative of 5/t^2 or 5t^-2=

5(-1t^-1)+ C= (-5)(t^-1) + C= -5/t + C, end result