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Integrate the function $\sqrt{16-x^2}-\frac{1}{8}\left(16-x^2\right)$ from 0 to $4$

Step-by-step Solution

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Final Answer

$7.233037$
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Step-by-step Solution

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1

Solve the product $-\frac{1}{8}\left(16-x^2\right)$

$\int_{0}^{4}\left(\sqrt{16-x^2}-2+\frac{1}{8}x^2\right)dx$

Learn how to solve definite integrals problems step by step online.

$\int_{0}^{4}\left(\sqrt{16-x^2}-2+\frac{1}{8}x^2\right)dx$

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Learn how to solve definite integrals problems step by step online. Integrate the function (16-x^2)^1/2-1/8(16-x^2) from 0 to 4. Solve the product -\frac{1}{8}\left(16-x^2\right). Expand the integral \int_{0}^{4}\left(\sqrt{16-x^2}-2+\frac{1}{8}x^2\right)dx into 3 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int_{0}^{4}\sqrt{16-x^2}dx results in: 4\pi . The integral \int_{0}^{4}-2dx results in: -8.

Final Answer

$7.233037$

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Function Plot

Plotting: $\sqrt{16-x^2}-\frac{1}{8}\left(16-x^2\right)$

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5
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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Definite Integrals

Given a function f(x) and the interval [a,b], the definite integral is equal to the area that is bounded by the graph of f(x), the x-axis and the vertical lines x=a and x=b

Used Formulas

4. See formulas

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