Mathematics
Senior High
Resolved
置き換える部分が違っていてもできるはずですがどこが間違っているのですか
4-7
dx
4-x= tεtic
4-x=1²
-1 = dt
4-t² = X
2t
dx
Date
dx=-2tdt.
x 03
t2→1
よって、
(4-1)².
X
25"
2
f
2
[
5
2dt
(16-812+1).dt
31
8
5
3
t
+16x
32
+
+16-(3-3-3-64
11
2x1
=-2x
462
5
24
R
-16
5
K
31
200
5
5
(2) 4-x=tとおく
与式
x=4-t
dx=(-1).dt
→
3
t471
(4-(-1)dt
√et
3
= 5,4 (16 t = - 8t² + t³ ³) dt
16
· [32x² - 11²²+ 3 t³]*
=
=32(2-1)-1/8(8-1)+/2/3(32-1)
= 32-172. 62. 106
+
15
Answers
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