For the following function f(x)f\left(x\right)f(x), find the antiderivative F(x)F\left(x\right)F(x) that satisfies the given condition.f(x)=5x4f\left(x\right)=5x^4f(x)=5x4; F(0)=1F\left(0\right)=1F(0)=1

F ( x ) = x 5 + 1 F\left(x\right)=x^5+1 F ( x ) = x 5 + 1

For the following function f ( x ) f\left(x\right) f ( x ) , find the antiderivative F ( x ) F\left(x\right) F ( x ) that satisfies the given condition. f ( x ) = 5 x 4 f\left(x\right)=5x^4 f ( x ) = 5 x 4 ; F ( 0 ) = 1 F\left(0\right)=1 F ( 0 ) = 1

the following function below, f of x, find the antiderivative big f of x that satisfies the given condition. So we have this function given to us, and we're asked to find the antiderivative. Now the first step I'm going to do is by finding this antiderivative function to give me the general solution. In doing this, well, I need to think about what I need to take the derivative of to give me 5x to the 4th. And if I think about it, that's gonna be x to the 5th, because I know by using this power rule, the 5 is gonna come out to the front, and then reducing this power by 1 will give me that 4 there. So that's going to be the function for the antiderivative, but we always have to add plus our constant c. So this here is the general case, the general solution where we have this plus c at the end here. We were trying to find the antiderivative that satisfies this condition. So that means we need to figure out what the c value is. And to do that, well, I can go ahead and plug this point in. So we have this function now. So we can see that f of 0 right here is equal to 1. So we're gonna have 0 to the 5th power plus c plugging in 0 for x, and we know that f of 0 is 1. So that means that 1 is going to equal 0 to the 5th power which is 0 plus c, meaning that c is equal to 1. So going back up to this function, we're going to have that our function f of x is equal to x to the 5th power plus this c value of 1. And that's going to be the solution for our function, and that is how you can solve this problem. Hope you found this video helpful, and let's move on.

For the following function f ( x ) f\left(x\right) f ( x ) , find the antiderivative F ( x ) F\left(x\right) F ( x ) that satisfies the given condition. f ( x ) = 5 x 4 f\left(x\right)=5x^4 f ( x ) = 5 x 4 ; F ( 0 ) = 1 F\left(0\right)=1 F ( 0 ) = 1

F ( x ) = x 5 + 1 F\left(x\right)=x^5+1 F ( x ) = x 5 + 1

F ( x ) = 5 x 5 + 1 F\left(x\right)=5x^5+1 F ( x ) = 5 x 5 + 1

F ( x ) = 25 x 5 + 1 F\left(x\right)=25x^5+1 F ( x ) = 25 x 5 + 1

F ( x ) = x 5 F\left(x\right)=x^5 F ( x ) = x 5

7. Antiderivatives & Indefinite Integrals

Antiderivatives

Business Calculus

7. Antiderivatives & Indefinite Integrals

Antiderivatives

Struggling with Business Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

For the following function $f\left(x\right)$ , find the antiderivative $F\left(x\right)$ that satisfies the given condition.
$f\left(x\right)=5x^4$ ; $F\left(0\right)=1$

$F\left(x\right)=5x^5+1$

$F\left(x\right)=x^5+1$

$F\left(x\right)=25x^5+1$

$F\left(x\right)=x^5$

Verified step by step guidance

Step 1: Recall that the antiderivative of a function f(x) is a function F(x) such that F'(x) = f(x). To find the antiderivative of f(x) = 5x^4, we use the power rule for integration.

Step 2: The power rule for integration states that for any term x^n, its antiderivative is (x^(n+1))/(n+1) + C, where C is the constant of integration. Apply this rule to f(x) = 5x^4.

Step 3: Integrate 5x^4. The exponent 4 increases by 1 to become 5, and the coefficient 5 is divided by the new exponent. This gives (5x^5)/5 + C, which simplifies to x^5 + C.

Step 4: Use the given condition F(0) = 1 to solve for the constant C. Substitute x = 0 into F(x) = x^5 + C, which gives 1 = 0^5 + C. Simplify to find C = 1.

Step 5: Substitute C = 1 back into the antiderivative to get the final function F(x) = x^5 + 1. This is the antiderivative that satisfies the given condition.

5:50m

Watch next

Master Antiderivatives with a bite sized video explanation from Patrick

Start learning

Antiderivatives practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice