Question

Find the most general antiderivative of the function. (Check your answer by differentiation.)

\[f\left( x \right) = 3\sqrt x - 2\sqrt[3]{x}\]

Short answer

The most general antiderivative of the function is \[F\left( x \right) = 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} - \frac{3}{2}{x^{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\]

Step-by-step solution

Theorem 1

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is\(F\left( x \right) + C\)where C is an arbitrary constant.

Antidifferentiation formula

• \(cf\left( x \right) = cF\left( x \right)\)

• \(f\left( x \right) + g\left( x \right) = F\left( x \right) + G\left( x \right)\)

• \({x^n}\left( {n \ne 1} \right) = \frac{{{x^{n + 1}}}}{{n + 1}}\)

Finding the most general antiderivative of the function

Given the function is,

\[f\left( x \right) = 3\sqrt x - 2\sqrt[3]{x}\]

Rewriting the function

\[f\left( x \right) = 3{x^{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}} - 2{x^{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}}\]

For finding antiderivative using Theorem 1 and the standard formula:

\[\begin{array}{l}F\left( x \right) = 3\left( {\frac{{{x^{\left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) + 1}}}}{{\left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) + 1}}} \right) - 2\left( {\frac{{{x^{\left( {{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}} \right) + 1}}}}{{\left( {{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}} \right) + 1}}} \right) + C\\F\left( x \right) = 3\left( {\frac{{{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}} \right) - 2\left( {\frac{{{x^{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}}} \right) + C\\F\left( x \right) = 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} - \frac{3}{2}{x^{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\end{array}\]

The most general antiderivative of the function is \[F\left( x \right) = 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} - \frac{3}{2}{x^{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\] .

Checking by differentiation

\[F\left( x \right) = 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} - \frac{3}{2}{x^{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\]

Differentiating with respect to x

\[\begin{array}{l}f\left( x \right) = 2\left( {\left( {{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}} \right){x^{\left( {{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) - 1}}} \right) - \frac{3}{2}\left( {\left( {{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}} \right){x^{\left( {{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}} \right) - 1}}} \right) + 0\\f\left( x \right) = 3{x^{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}} - 2{x^{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}}\\f\left( x \right) = 3\sqrt x - 2\sqrt[3]{x}\end{array}\]

Since the original function is obtained so antiderivative is correct.