Trabajo: Cálculo de integrales indefinidas, no inmediatas
Enviado por esteban2602 • 16 de Abril de 2023 • Trabajo • 815 Palabras (4 Páginas) • 41 Visitas
𝑏
𝑣 = ∫ 𝐴(𝑥)𝑑𝑥 → 𝑀𝐸𝑇𝑂𝐷𝑂 𝐷𝐸 𝐷𝐼𝑆𝐶𝑂𝑆
𝑎
Funciones:
[pic 1]
𝐴(𝑥) = 𝜋[𝑓(𝑥)]2 − 𝜋[𝑔(𝑥)]2 = 𝜋([𝑓(𝑥)]2 − [𝑔(𝑥)]2)
𝐴(𝑥) = 𝜋([2𝑥 𝑒
−𝑥]2
𝑥
− [𝑠𝑒𝑛 ([pic 2]
2
)] )
10
𝐴(𝑥) = 𝜋(4𝑥 𝑒−2𝑥 2 𝑥 )) 10[pic 3]
𝑉 = 𝜋 ∫ (4𝑥2 𝑒−2𝑥 − 𝑠𝑒𝑛2( 𝑥[pic 4][pic 5]
0 10
Definición de integración por partes:
))𝑑𝑥
∫ 𝑢𝑣1 = 𝑢𝑣 − ∫ 𝑢1𝑣
𝑢 = 𝑥2 𝑑 𝑢1 = (𝑥2) = 2𝑥 𝑑𝑥 | 𝑣1 = 𝑒−2𝑥 𝑣 = ∫ 𝑒−2𝑥𝑑𝑥 = − 1 𝑒−2𝑥 2 |
= [𝑥2 (− 1[pic 6][pic 7][pic 8]
2
𝑒−2𝑥) − ∫ 2𝑥 (− 1
2[pic 9]
2
𝑒−2𝑥) 𝑑𝑥]
0
Simplifico:
1
= 4 [−[pic 10]
2
2
𝑒−2𝑥𝑥2 − ∫ −𝑒−2𝑥𝑥𝑑𝑥]
0
= 4 [−
1 𝑒−2𝑥𝑥2 − (− 1
2 4[pic 11][pic 12]
2
(−2𝑒−2𝑥𝑥 − 𝑒−2𝑥))]
0
= 4 [−
1 𝑒−2𝑥𝑥2 +
2[pic 13]
1 2
(−2𝑒−2𝑥𝑥 − 𝑒−2𝑥)] 4 0[pic 14]
Calculo limites:
lim → 0 + (− 1 𝑒−2𝑥𝑥2 + 1 (−2𝑒−2𝑥𝑥 − 𝑒−2𝑥))[pic 15][pic 16]
𝑥 2 4
= − 1 𝑒−2×0 × 02 + 1 (−2𝑒−2×0 × 0 − 𝑒−2×0) = − 1[pic 17][pic 18][pic 19]
2 4 4
lim → 2 + (− 1 𝑒−2𝑥𝑥2 + 1 (−2𝑒−2𝑥𝑥 − 𝑒−2𝑥))[pic 20][pic 21]
𝑥 2 4
= − 1 𝑒−2×2 × 22 + 1 (−2𝑒−2×2 × 2 − 𝑒−2×2) = − 13[pic 22][pic 23][pic 24]
2 4 4𝑒4
13 1 13
= 4 (− 4𝑒4 + 4) = − 𝑒4 + 1[pic 25][pic 26][pic 27]
...