find the area under the curve y=4x^2 from x=0 to x=5
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Step-by-step explanation:
First, we need to find where the two functions intersect.
4
x
−
x
2
=
x
x
2
−
4
x
+
x
=
0
x
2
−
3
x
=
0
x
(
x
−
3
)
=
0
x
=
0
and
x
=
3
So our interval of integration is
0
≤
x
≤
3
y
=
4
x
−
x
2
is a parabola that is concave down
y
=
x
is the line that passes through the origin with slope 1
The integral for the area is
∫
4
x
−
x
2
−
x
d
x
=
∫
3
x
−
x
2
d
x
integrating we have
3
2
x
2
−
1
3
x
3
Evaluating at we have
(
3
2
)
3
2
−
(
1
3
)
3
3
−
0
=
27
2
−
27
3
=
27
2
−
9
81
6
−
54
6
=
27
6
=
9
2
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