Consider the function f(x)=x^4+7sqrtx

Question

Consider the function f(x)=x^4+7sqrtx

Let F(x) be the antiderivative of f(x) with F(1)=−2.

Then F(x)= _______

Answer 1

The antiderivative is the function that, if you take its derivative, you get f(x).. So, what would that function look like?

Well, if you take the derivative and get x⁴, it must be some form of x⁵ since the power rule reduces the exponent by one. But if it were x⁵, the derivative would be 5x⁴. To fix this, start with (1/5)x⁵ then the derivative will work out. The same goes for 7√x or 7x^1/2, you would need to start with some form of x^3/2 so, reduced by one, the exponent would become 1/2 when you take the derivative. But x^3/2 would have a derivative of (3/2)x^1/2, to fix that, start with (14/3)x^3/2 and the derivative will work out just right.

So the antiderivative is F(x) = (1/5)x⁵ + (14/3)x^3/2 + C where "C" is any constant (because the derivative of a constant is zero. To determine what the correct value of "C" is, use the given fact that F(1) = -2

-2 = (1/5)(1)⁵ + (14/3)(1)^3/2 + C

-2 = 1/5 + 14/3 + C

-30/15 = 3/15 + 70/15 + C

C = -30/15 - 3/15 - 70/15 = -103/15

F(x) = (1/5)x⁵ + (14/3)x^3/2 - 103/15

Answer 2

To find the antiderivative, for each term raise the exponent by 1, then multiply by the reciprocal of the new exponent.

For x⁴, the antiderivative is 1/5 x⁵.

For 7x^1/2, the antiderivative is 7(2/3)x^3/2 or 14/3 x^3/2.

So the general antiderivative is 1/5 x⁵ + 14/3 x^3/2 +C

To find the value of C, when x = 1, F(x) = -2.

-2 = 1/5(1) + 14/3 (1) + C

C = -103/15

So, F(x) = 1/5 x⁵ + 14/3 x^3/2-103/15

Notice of this graph that the graphs of f(x) and F'(x) are identical.

desmos.com/calculator/guquipocvw

Answer 3

F(x) = x⁵/5 + 14/3x^3/2 + C; F(1) = 1/5 + 14/3 + C = - 2; C = - 103/15; F(x) = x⁵/5 + 14/3x^3/2 - 103/15