Question

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

\(f\left( x \right) = 7{x^{{2 \mathord{\left/

{\vphantom {2 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 8{x^{{{ - 4} \mathord{\left/

{\vphantom {{ - 4} 5}} \right.

\kern-\nulldelimiterspace} 5}}}\)

Short answer

The most general antiderivative of the function is \(F\left( x \right) = 5{x^{{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 40{x^{{1 \mathord{\left/

{\vphantom {1 5}} \right.

\kern-\nulldelimiterspace} 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

Given the function is,

\(f\left( x \right) = 7{x^{{2 \mathord{\left/

{\vphantom {2 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 8{x^{{{ - 4} \mathord{\left/

{\vphantom {{ - 4} 5}} \right.

\kern-\nulldelimiterspace} 5}}}\)

For finding antiderivative using Theorem 1 and the standard formula:

\(\begin{aligned}{l}F\left( x \right) = 7\left( {\frac{{{x^{\left( {{2 \mathord{\left/

{\vphantom {2 5}} \right.

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

{\vphantom {2 5}} \right.

\kern-\nulldelimiterspace} 5}} \right) + 1}}} \right) + 8\left( {\frac{{{x^{\left( {{{ - 4} \mathord{\left/

{\vphantom {{ - 4} 5}} \right.

\kern-\nulldelimiterspace} 5}} \right) + 1}}}}{{\left( {{{ - 4} \mathord{\left/

{\vphantom {{ - 4} 5}} \right.

\kern-\nulldelimiterspace} 5}} \right) + 1}}} \right) + C\\F\left( x \right) = 7\left( {\frac{{{x^{{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}}}}}{{{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}}}} \right) + 8\left( {\frac{{{x^{{1 \mathord{\left/

{\vphantom {1 5}} \right.

\kern-\nulldelimiterspace} 5}}}}}{{{1 \mathord{\left/

{\vphantom {1 5}} \right.

\kern-\nulldelimiterspace} 5}}}} \right) + C\\F\left( x \right) = 5{x^{{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 40{x^{{1 \mathord{\left/

{\vphantom {1 5}} \right.

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

The most general antiderivative of the function is \(F\left( x \right) = 5{x^{{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 40{x^{{1 \mathord{\left/

{\vphantom {1 5}} \right.

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

Checking by differentiation

\(F\left( x \right) = 5{x^{{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 40{x^{{1 \mathord{\left/

{\vphantom {1 5}} \right.

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

Differentiating with respect to x.

\(\begin{aligned}{l}f\left( x \right) = 5\left( {\frac{7}{5}{x^{\left( {{7 \mathord{\left/

{\vphantom {7 5}} \right.

\kern-\nulldelimiterspace} 5}} \right) - 1}}} \right) + 40\left( {\frac{1}{5}{x^{\left( {{1 \mathord{\left/

{\vphantom {1 5}} \right.

\kern-\nulldelimiterspace} 5}} \right) - 1}}} \right) + 0\\f\left( x \right) = 7{x^{{2 \mathord{\left/

{\vphantom {2 5}} \right.

\kern-\nulldelimiterspace} 5}}} + 8{x^{{{ - 4} \mathord{\left/

{\vphantom {{ - 4} 5}} \right.

\kern-\nulldelimiterspace} 5}}}\end{aligned}\)

Since the original function is obtained so antiderivative is correct.