∫ 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 |
onsdag den 1. juli 2015
Integral-bestemmelse (eksempler)
Abonner på:
Kommentarer til indlægget (Atom)
Ingen kommentarer:
Send en kommentar