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Math  /  Calculus

QuestionF(x)=F(x)=

Studdy Solution

STEP 1

1. We are given the function f(x)=6x3−7x23 f(x) = \frac{6}{\sqrt[3]{x}} - 7\sqrt[3]{x^2} .
2. We need to find the most general antiderivative of this function.
3. We will use an upper-case C C to represent the constant of integration.

STEP 2

1. Rewrite the function in terms of powers of x x .
2. Integrate each term separately.
3. Combine the integrated terms and add the constant of integration C C .

STEP 3

Rewrite f(x)=6x3−7x23 f(x) = \frac{6}{\sqrt[3]{x}} - 7\sqrt[3]{x^2} using exponents:
f(x)=6x−13−7x23 f(x) = 6x^{-\frac{1}{3}} - 7x^{\frac{2}{3}}

STEP 4

Integrate the first term 6x−13 6x^{-\frac{1}{3}} :
∫6x−13 dx=6⋅x−13+1−13+1=6⋅x2323=6⋅32x23=9x23 \int 6x^{-\frac{1}{3}} \, dx = 6 \cdot \frac{x^{-\frac{1}{3} + 1}}{-\frac{1}{3} + 1} = 6 \cdot \frac{x^{\frac{2}{3}}}{\frac{2}{3}} = 6 \cdot \frac{3}{2} x^{\frac{2}{3}} = 9x^{\frac{2}{3}}

STEP 5

Integrate the second term −7x23 -7x^{\frac{2}{3}} :
∫−7x23 dx=−7⋅x23+123+1=−7⋅x5353=−7⋅35x53=−215x53 \int -7x^{\frac{2}{3}} \, dx = -7 \cdot \frac{x^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} = -7 \cdot \frac{x^{\frac{5}{3}}}{\frac{5}{3}} = -7 \cdot \frac{3}{5} x^{\frac{5}{3}} = -\frac{21}{5}x^{\frac{5}{3}}

STEP 6

Combine the results of the integrations and add the constant of integration C C :
F(x)=9x23−215x53+C F(x) = 9x^{\frac{2}{3}} - \frac{21}{5}x^{\frac{5}{3}} + C
The most general antiderivative of the function is:
F(x)=9x23−215x53+C F(x) = 9x^{\frac{2}{3}} - \frac{21}{5}x^{\frac{5}{3}} + C

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