Math, asked by pravinkumarpathak1, 2 months ago

Solve this,,,,,

1) ∫x³.dx

2) ∫x⁹.dx

3) ∫x².dx

Answers

Answered by Anonymous
18

1)

\int {x}^{3} dx

 = >  \frac{ {x}^{3 + 1} }{3 + 1}  + c

 =  >   \frac{ {x}^{4} }{4}

2)

\int {x}^{9} .dx

 = >  \frac{ {x}^{9 + 1} }{9 + 1}  + c

 = >  \frac{ {x}^{10} }{10}

3)

\int \:  {x}^{2} .dx

 =  >  \frac{ {x}^{2 + 1} }{2 + 1}

 = >  \frac{ {x}^{3} }{3}

Answered by Anonymous
93

Explanation :

According to the power rule,

\bold \red{ \int x {}^{n} .dx}  \:  \rightarrow \bold \green{ \dfrac{x {}^{n + 1} }{n + 1}  + c}

Case (I),  \sf \int x {}^{3} .dx

 \mapsto \sf \:  \dfrac{x {}^{3 + 1} }{3 + 1}  + c

 \mapsto { \boxed{ \sf{\dfrac{x {}^{4} }{4}  + c}}} \:  \bigstar

ㅤㅤㅤㅤㅤ

Case (II), \sf \int x {}^{9} .dx

 \mapsto \sf  \:  \dfrac{x {}^{9 + 1} }{9 + 1}  + c

\mapsto{ \boxed{ \sf{ \frac{x {}^{10} }{10}  + c}}} \:  \bigstar

ㅤㅤㅤㅤㅤ

Case (III), \sf\int x {}^{2} .dx

\mapsto \sf  \:  \dfrac{x {}^{2 + 1} }{2 + 1}  + c

\mapsto{ \boxed{ \sf{ \frac{x {}^{3} }{3}  + c}}} \:  \bigstar

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