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Math  /  Calculus

QuestionDecide which function is an antiderivative of the other: f(x)=5xg(x)=10xf(x)=\frac{5}{\sqrt{x}} \quad g(x)=10 \sqrt{x}
Choose one

Studdy Solution

STEP 1

1. We are given two functions: f(x)=5x f(x) = \frac{5}{\sqrt{x}} and g(x)=10x g(x) = 10 \sqrt{x} .
2. We need to determine which function is an antiderivative of the other.

STEP 2

1. Find the antiderivative of f(x) f(x) .
2. Find the antiderivative of g(x) g(x) .
3. Compare the results to determine which function is an antiderivative of the other.

STEP 3

Find the antiderivative of f(x)=5x f(x) = \frac{5}{\sqrt{x}} .
Rewrite f(x) f(x) in terms of exponents:
f(x)=5x−1/2 f(x) = 5x^{-1/2}

STEP 4

Integrate f(x)=5x−1/2 f(x) = 5x^{-1/2} :
∫5x−1/2 dx=5⋅x1/21/2+C \int 5x^{-1/2} \, dx = 5 \cdot \frac{x^{1/2}}{1/2} + C
Simplify:
=10x1/2+C = 10x^{1/2} + C

STEP 5

Find the antiderivative of g(x)=10x g(x) = 10\sqrt{x} .
Rewrite g(x) g(x) in terms of exponents:
g(x)=10x1/2 g(x) = 10x^{1/2}

STEP 6

Integrate g(x)=10x1/2 g(x) = 10x^{1/2} :
∫10x1/2 dx=10⋅x3/23/2+C \int 10x^{1/2} \, dx = 10 \cdot \frac{x^{3/2}}{3/2} + C
Simplify:
=203x3/2+C = \frac{20}{3}x^{3/2} + C

STEP 7

Compare the antiderivatives:
The antiderivative of f(x)=5x f(x) = \frac{5}{\sqrt{x}} is 10x+C 10\sqrt{x} + C , which matches g(x) g(x) .
The antiderivative of g(x)=10x g(x) = 10\sqrt{x} is 203x3/2+C \frac{20}{3}x^{3/2} + C , which does not match f(x) f(x) .
Therefore, g(x)=10x g(x) = 10\sqrt{x} is an antiderivative of f(x)=5x f(x) = \frac{5}{\sqrt{x}} .
The function g(x)=10x g(x) = 10\sqrt{x} is an antiderivative of f(x)=5x f(x) = \frac{5}{\sqrt{x}} .

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