Question

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

\(f\left( x \right) = \sqrt(3){{{x^2}}} + x\sqrt x \)

Short answer

The most general antiderivative of the function is \(F\left( x \right) = \frac{{3{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{5} + \frac{{2{x^{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{5} + 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

Step 3: Finding the most general antiderivative of the function

Given the function is,

\(f\left( x \right) = \sqrt(3){{{x^2}}} + x\sqrt x \)

Rewriting the function:

\(\begin{aligned}{l}f\left( x \right) = {x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}} + {x^{1 + \left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

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

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}} + {x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}\end{aligned}\)

For finding antiderivative using Theorem 1 and the standard formula:

\(\begin{aligned}{l}F\left( x \right) = \frac{{{x^{\left( {{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}} \right) + 1}}}}{{{2 \mathord{\left/

{\vphantom {2 {3 + 1}}} \right.

\kern-\nulldelimiterspace} {3 + 1}}}} + \frac{{{x^{\left( {{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) + 1}}}}{{{3 \mathord{\left/

{\vphantom {3 {2 + 1}}} \right.

\kern-\nulldelimiterspace} {2 + 1}}}} + C\\F\left( x \right) = \left( {\frac{{{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}} \right) + \left( {\frac{{{x^{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}}} \right) + C\\F\left( x \right) = \frac{{3{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{5} + \frac{{2{x^{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{5} + C\end{aligned}\)

The most general antiderivative of the function is \(F\left( x \right) = \frac{{3{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{5} + \frac{{2{x^{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{5} + C\).

Checking by differentiation

\(F\left( x \right) = \frac{{3{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{5} + \frac{{2{x^{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{5} + C\)

Differentiating with respect to x:

\(\begin{aligned}{l}f\left( x \right) = \frac{{3\left( {\left( {{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}} \right){x^{\left( {{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}} \right) - 1}}} \right)}}{5} + \frac{{2\left( {\left( {{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}} \right){x^{\left( {{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) - 1}}} \right)}}{5} + 0\\f\left( x \right) = {x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}} + {x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}\\f\left( x \right) = \sqrt(3){{{x^2}}} + x\sqrt x \end{aligned}\)

Since the original function is obtained so antiderivative is correct.