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Integrate the function $-161+x^2$ from 0 to $2$

Step-by-step Solution

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Final answer to the problem

$-\frac{958}{3}$
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Step-by-step Solution

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  • Product of Binomials with Common Term
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Expand the integral $\int_{0}^{2}\left(-161+x^2\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int_{0}^{2}-161dx+\int_{0}^{2} x^2dx$

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

$\int_{0}^{2}-161dx+\int_{0}^{2} x^2dx$

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Learn how to solve definite integrals problems step by step online. Integrate the function -161+x^2 from 0 to 2. Expand the integral \int_{0}^{2}\left(-161+x^2\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int_{0}^{2}-161dx results in: -322. The integral \int_{0}^{2} x^2dx results in: \frac{8}{3}. Gather the results of all integrals.

Final answer to the problem

$-\frac{958}{3}$

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Function Plot

Plotting: $-161+x^2$

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5
6
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

See formulas (3)

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