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)=100x99f\left(x\right)=100x^{99}f(x)=100x99; F(1)=101F\left(1\right)=101F(1)=101

F(x)=x100+100F\left(x\right)=$$x^{100}+100F(x)=x100+100$$

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 ) = 100 x 99 f\left(x\right)=100x^{99} f ( x ) = 100 x 99 ; F ( 1 ) = 101 F\left(1\right)=101 F ( 1 ) = 101

this problem, we are given the following function f of x, and we're asked to find the antiderivative, big f of x, that satisfies the given condition. So we have this function, we have our given condition. And to start things off, what I'm going to do is see if I can find the general case for the antiderivative. So doing that is going to be finding big f of x, and we'll find this general solution by thinking about what derivative would we have had to take in order to get this function, a 100x to the 99th power. Now this might look like this really kind of big complicated function, but notice we have this big number out in front, and then we have this exponent which is one less than that number. It kinda looks like we used the power rule here of some kind. Well, if I had x to the 100th power, this number out in front here, we could take that number, multiply it by the front. That would give us the 100 out front, and then reducing the power by 1 would give us 99. So this derivative would give us the original function we had. So this would be the antiderivative of our function, but you always have to add a plus c when you take this antiderivative. So this would be the general solution. But I'm not looking for the general case where we have plus c in this equation. I'm looking for the specific or particular case where we need to find what big f of x is that satisfies this condition. So to do that, well, I'm gonna plug my numbers in. So I can see that I'm plugging in f of 1. So to find f of 1, we take every place that we see x and we replace it with 1. So we only have 1 to the 100th power plus c. Now f of 1 is given to us here as a 101. So we have a 101 is equal to 1 to the 100th power, which 1 multiplied by itself a 100 times is still just gonna be 1. Because 1 multiplied by 1 is would just be 1. And we do that a 100 times, we still get 1 plus c. And from here, I'm just gonna subtract this 1 on both sides of the equation that gets the ones to cancel on the right side here, giving me that c is equal to a 101 minus 1, which is a 100. So what that means is that our function is going to be big f of x is equal to x to the 100th power plus c, which in this case is a 100. So this right here would be our function big f of x, and that is the particular solution to this problem. So that is how you can solve these situations where you need to find an antiderivative and find a particular solution. Hope you found this video helpful.

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)=100x99f\left(x\right)=$$100x^{99}f(x)=100x99$$; F(1)=101F\left(1\right)=101F(1)=101

F(x)=x100+100F\left(x\right)=$$x^{100}+100F(x)=x100+100$$

F(x)=x100+101F\left(x\right)=$$x^{100}+101F(x)=x100+101$$

F(x)=100x100+101F\left(x\right)=$$100x^{100}+101F(x)=100x100+101$$

F(x)=100x100+1F\left(x\right)=$$100x^{100}+1F(x)=100x100+1$$

7. Antiderivatives & Indefinite Integrals

Antiderivatives

Business Calculus

7. Antiderivatives & Indefinite Integrals

Antiderivatives

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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$)=100x^{99}$ ; $F$\left$(1$\right$)=101$

$F$\left$(x$\right$)=x^{100}+101$

$F$\left$(x$\right$)=x^{100}+100$

$F$\left$(x$\right$)=100x^{100}+101$

$F$\left$(x$\right$)=100x^{100}+1$

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) = 100x^99, we will use the power rule for integration.

Step 2: The power rule for integration states that for any term x^n, the antiderivative is (x^(n+1))/(n+1) + C, where C is the constant of integration. Apply this rule to f(x) = 100x^99.

Step 3: Integrate 100x^99. Increase the exponent by 1 (from 99 to 100) and divide the coefficient (100) by the new exponent (100). This gives the term (100x^100)/100, which simplifies to x^100. Add the constant of integration, C, to get F(x) = x^100 + C.

Step 4: Use the given condition F(1) = 101 to solve for the constant C. Substitute x = 1 into F(x) = x^100 + C, which gives 101 = (1)^100 + C. Simplify to find C = 101 - 1 = 100.

Step 5: Substitute the value of C back into the antiderivative. The final expression for F(x) is F(x) = x^100 + 100.

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Struggling with Business Calculus?

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)=100x99f\(\left\)(x\(\right\))=100x^{99}f(x)=100x99; F(1)=101F\(\left\)(1\(\right\))=101F(1)=101

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For the following function $f\(\left\)(x\(\right\))$ , find the antiderivative $F\(\left\)(x\(\right\))$ that satisfies the given condition.
$f\(\left\)(x\(\right\))=100x^{99}$ ; $F\(\left\)(1\(\right\))=101$