onsdag den 1. juli 2015

Integral-bestemmelse (eksempler)


∫ 4e2x dx = 

4 ⋅∫ e2x dx =

4 ⋅ 12 ⋅ e2x + k =

2 ⋅ e2x + k




∫ ( x6 + 1x ) dx =

17 ⋅ x7 + ln ( x ) + k, x > 0




∫− 12√ x dx =

−√ x + k, x > 0




∫( 35 ⋅ x5 + x3 − x − 5 ) dx = 

3 ⋅ x65 ⋅ 6 + 14 ⋅ x4 − 12 ⋅ x2 − 5x + k


x610 + x44 − x22 − 5x + k





∫( 17 − 4x + e−5x ) dx =

17x − 4 ⋅ ln ( x ) − e−5x5 + k, x > 0

∫( 6x − 3e−x ) dx =

6 ⋅ 12 ⋅ x2 − 3 ⋅ 1( −1 ) ⋅ e−x + k =

3 ⋅ x2 + 3 ⋅ e−x + k =


3 ⋅ ( x2 + e−x ) + k =





Fra afledt funktion til stamfunktion

















∫( x1,5 − 2 + 8x2 ) dx =

x2,52,5 − 2x + 8 ⋅∫ ( x0x2 ) dx =


x2,52,5 − 2x + 8 ⋅∫ ( x−2 ) dx =

x2,52,5 − 2x + 8 ⋅ x−2 + 1 −1 + k =

x2,52,5 − 2x + 8 ⋅ ( −x−1 ) + k =

x2,52,5 − 2x − 8x

 + k




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