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Integrate the function $4-x^2-y$ from 0 to $1$

Step-by-step Solution

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Final Answer

$\frac{11}{3}-y$
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Step-by-step Solution

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1

Expand the integral $\int_{0}^{1}\left(4-x^2-y\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$\int_{0}^{1}4dx+\int_{0}^{1}-x^2dx+\int_{0}^{1}-ydx$

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

$\int_{0}^{1}4dx+\int_{0}^{1}-x^2dx+\int_{0}^{1}-ydx$

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

Final Answer

$\frac{11}{3}-y$

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Function Plot

Plotting: $4-x^2-y$

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1
2
3
4
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

3. See formulas

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