integration (xz/(16-x^2)
Answers
Answer:
Although it is a possibility, a trigonometric substitution is not necessary.
This can also be tackled using the substitution
u
=
16
−
x
2
. This implies that
d
u
=
−
2
x
d
x
. Thus:
∫
4
0
x
√
16
−
x
2
d
x
=
−
1
2
∫
4
0
−
2
x
√
16
−
x
2
d
x
Before making the
u
and
d
u
substitutions, recall that the bounds will change. Plug the current bounds into
16
−
x
2
. Thus the bound of
0
becomes
16
−
0
2
=
16
and the bound of
4
becomes
16
−
4
2
=
0
.
−
1
2
∫
4
0
−
2
x
√
16
−
x
2
d
x
=
−
1
2
∫
0
16
1
√
u
d
u
From here, we can reorder the integral using the rule:
∫
b
a
f
(
x
)
d
x
=
−
∫
a
b
f
(
x
)
d
x
. Also, rewrite
1
√
u
using fractional and negative exponents:
−
1
2
∫
0
16
1
√
u
d
u
=
1
2
∫
16
0
u
−
1
2
d
u
From here, integrate using the rule:
∫
u
n
d
u
=
u
n
+
1
n
+
1
and then evaluate the integral.
1
2
∫
16
0
u
−
1
2
d
u
=
1
2
[
u
−
1
2
+
1
−
1
2
+
1
]
16
0
=
1
2
[
u
1
2
1
2
]
16
0
The
1
2
can be brought from the denominator as a
2
, cancelling with the
1
2
lingering outside of the brackets, leaving:
1
2
[
u
1
2
1
2
]
16
0
=
[
√
u
]
16
0
=
√
16
−
√
0
=
4
Answer:
Step-by-step explanation:
X