y2 | ||
∫−12(xy+ | |0x2+3)dx= | |
2 |
(x2+3)2 | 02 | |||
∫−12{x(x2+3)+ | −x*0− | )dx = | ||
2 | 2 |
x4+6x2+9 | ||
∫−12(x3+3x + | )dx= ............. | |
2 |
1 | ||
=−1∫2([xy+ | y2]0x2+3)dx= | |
2 |
1 | ||
=−1∫2[x*(x2+3)+ | *(x2+3)2−0]dx= | |
2 |
1 | 9 | |||
=−1∫2[x3+3x+ | x4+3x2+ | ] dx= | ||
2 | 2 |
1 | 1 | 1 | 1 | 1 | 9 | |||||||
=[ | x4+3* | x2+ | * | x5+3* | x3+ | x]−12= | ||||||
4 | 2 | 2 | 5 | 3 | 2 |
1 | 1 | 3 | 9 | |||||
=[ | x5+ | x4+x3+ | x2+ | x]−12= | ||||
10 | 4 | 2 | 2 |
1 | 1 | 3 | 1 | 1 | 3 | 9 | ||||||||
= | *32+ | *16+8+ | *4+9−(− | + | −1+ | − | )= | |||||||
10 | 4 | 2 | 10 | 4 | 2 | 2 |
32 | 1 | 1 | ||||
= | +27−(−4 | + | )= | |||
10 | 10 | 4 |
681 | ||
= | ||
20 |