Primitivna funkcija

Enostavna integracija

Integrirati a x n {\displaystyle ax^{n}}

• Prištejte 1 k številu n {\displaystyle n} tako da je a x n {\displaystyle ax^{n}} zdaj a x n + 1 {\displaystyle ax^{n+1}}
• Vse to delimo z novo močjo, tako da je zdaj a x n + 1 n + 1 {\displaystyle {\frac {ax^{n+1}}{n+1}}}
• Dodamo konstanto c {\displaystyle c} , tako da je zdaj a x n + 1 n + 1 + c {\displaystyle {\frac {ax^{n+1}}}{n+1}}+c}

To je mogoče prikazati kot:

∫ a x n d x = a x n + 1 n + 1 + c {\displaystyle \int ax^{n}\ dx={\frac {ax^{n+1}}}{n+1}}+c}

Kadar je veliko x {\displaystyle x} členov, integrirajte vsak del posebej:

∫ 2 x 6 - 5 x 4 d x = 2 x 7 7 - 5 x 5 5 + c = 2 7 x 7 - x 5 + c {\displaystyle \int 2x^{6}-5x^{4}\ dx={\frac {2x^{7}}{7}}-{\frac {5x^{5}}{5}}+c={\frac {2}{7}}x^{7}-x^{5}+c}

(To deluje le, če se deli dodajajo ali odvzemajo.)

Primeri

∫ 3 x 4 d x = 3 x 5 5 + c {\displaystyle \int 3x^{4}\ dx={\frac {3x^{5}}{5}}+c}

∫ x + x 2 + x 3 + x 4 d x = x 2 2 + x 3 3 + x 4 4 + x 5 5 + c {\displaystyle \int x+x^{2}+x^{3}+x^{4}\ dx={\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}}+c}

∫ 1 x + 4 d x = ln | x + 4 | × 1 + c = ln | x + 4 | + c {\displaystyle \int {\frac {1}{x+4}} dx=\ln |x+4|krat 1+c=\ln |x+4|+c}

Preoblikovanje ulomkov in korenov v moči je lažje:

∫ 1 x 3 d x = ∫ x - 3 d x = x - 2 - 2 + c = - 1 2 x 2 + c {\displaystyle \int {\frac {1}{x^{3}}}}\x=\int x^{-3}\ dx={\frac {x^{-2}}{-2}}+c=-{\frac {1}{2x^{2}}}+c}

∫ x 3 d x = ∫ x 3 2 d x = x 5 2 5 2 + c = 2 5 x 5 2 + c = 2 5 x 5 + c {\displaystyle \int {\sqrt {x^{3}}} dx=\int x^{\frac {3}{2}}\ dx={\frac {x^{\frac {5}{2}}}{\frac {5}{2}}}+c={\frac {2}{5}}x^{\frac {5}{2}}+c={\frac {2}{5}}{\sqrt {x^{5}}}+c}

Integracija oklepaja ("verižno pravilo")

Če želite integrirati oklepaj, kot je ( 2 x + 4 ) 3 {\displaystyle (2x+4)^{3}} , moramo to storiti na drugačen način. Imenuje se verižno pravilo. To je podobno preprostemu integriranju. Deluje le, če ima x {\displaystyle x} v oklepaju moč 1 (je linearen), na primer x {\displaystyle x} ali 5 x {\displaystyle 5x} (ne x 5 {\displaystyle x^{5}} ali x - 7 {\displaystyle x^{-7}} ).

Za izvedbo ∫ ( 2 x + 4 ) 3 d x {\displaystyle \int (2x+4)^{3}\ dx}

• Dodamo 1 na moč 3 {\displaystyle 3} , tako da je zdaj ( 2 x + 4 ) 4 {\displaystyle (2x+4)^{4}}
• Vse to delimo z novo močjo in dobimo ( 2 x + 4 ) 4 4 4 {\displaystyle {\frac {(2x+4)^{4}}{4}}}}
• Vse to delimo z derivativom oklepaja ( d ( 2 x + 4 ) d x = 2 ) {\displaystyle \left({\frac {d(2x+4)}{dx}}=2\right)} da dobimo ( 2 x + 4 ) 4 4 4 × 2 = 1 8 ( 2 x + 4 ) 4 {\displaystyle {\frac {(2x+4)^{4}}{4\times 2}}={\frac {1}{8}}(2x+4)^{4}}
• Dodamo konstanto c {\displaystyle c} in dobimo 1 8 ( 2 x + 4 ) 4 + c {\displaystyle {\frac {1}{8}}(2x+4)^{4}+c}

Primeri

∫ ( x + 1 ) 5 d x = ( x + 1 ) 6 6 × 1 + c = 1 6 ( x + 1 ) 6 + c ( ∵ d ( x + 1 ) d x = 1 ) {\displaystyle \int (x+1)^{5}\ dx={\frac {(x+1)^{6}}{6\krat 1}}+c={\frac {1}{6}}(x+1)^{6}+c\left(\brez {\frac {d(x+1)}{dx}}=1\right)}

∫ 1 ( 7 x + 12 ) 9 d x = ∫ ( 7 x + 12 ) - 9 d x = ( 7 x + 12 ) - 8 - 8 × 7 + c = - 1 56 ( 7 x + 12 ) - 8 + c = - 1 56 ( 7 x + 12 ) 8 + c ( ∵ d ( 7 x + 12 ) d x = 7 ) {\displaystyle \int {\frac {1}{(7x+12)^{9}}} dx=\int (7x+12)^{-9}\ dx={\frac {(7x+12)^{-8}}{-8krat 7}}+c=-{\frac {1}{56}}(7x+12)^{-8}+c=-{\frac {1}{56(7x+12)^{8}}}}+c\left(\because {\frac {d(7x+12)}{dx}}=7\right)}

Sorodne strani

• Matematika
• Integralno