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Integral
Integral Solve for A and 2A where A = Area of a triangle.
∫
2
e
+1
(
ln
(
x
)
+
ln
(
1
−
x
1
)
/
(
x
−
1
)
)
d
x
=
A
Please graph above Integral to determine the Area within the shape. Below is an incorrect example of such graph.
–2
–2
–2
–1
–1
–1
1
1
1
2
2
2
–1
–1
–1
1
1
1
0
0
0
f
f
f
c
c
c
1
1
1
= 0
= 0
= 0
Graph
–2
–2
–2
–1
–1
–1
1
1
1
2
2
2
–1
–1
–1
1
1
1
0
0
0
Graph
–2
–2
–2
–1
–1
–1
1
1
1
2
2
2
3
3
3
1
1
1
2
2
2
0
0
0
f
f
f
c
c
c
1
1
1
= 1.8
= 1.8
= 1.8
c
c
c
2
2
2
= 4
= 4
= 4
B
B
B
Graph
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