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Integrate the function $-11\sqrt{16+x^2}$ from $-41$ to $1$

Step-by-step Solution

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Final Answer

$9599.855971$
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Step-by-step Solution

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The integral of a constant times a function is equal to the constant multiplied by the integral of the function

$-11\int_{-41}^{1}\sqrt{16+x^2}dx$

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$-11\int_{-41}^{1}\sqrt{16+x^2}dx$

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Learn how to solve definite integrals problems step by step online. Integrate the function -11(16+x^2)^1/2 from -41 to 1. The integral of a constant times a function is equal to the constant multiplied by the integral of the function. We can solve the integral \int\sqrt{16+x^2}dx by applying integration method of trigonometric substitution using the substitution. Now, in order to rewrite d\theta in terms of dx, we need to find the derivative of x. We need to calculate dx, we can do that by deriving the equation above. Substituting in the original integral, we get.

Final Answer

$9599.855971$

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Function Plot

Plotting: $-11\sqrt{16+x^2}$

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1
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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

8. See formulas

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