Integrate 4-1x^2 from -2 to 2

\int_{-2}^{2}\left(4-x^2\right)dx

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

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

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Answer

$\frac{32}{3}$

Step by step solution

Problem

$\int_{-2}^{2}\left(4-x^2\right)dx$
1

The integral of a sum of two or more functions is equal to the sum of their integrals

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

The integral of a constant is equal to the constant times the integral's variable

$\int_{-2}^{2}-x^2dx+\left[4x\right]_{-2}^{2}$
3

Evaluate the definite integral

$\int_{-2}^{2}-x^2dx-1\left(-2\right)\cdot 4+2\cdot 4$
4

Multiply $4$ times $2$

$\int_{-2}^{2}-x^2dx+8+8$
5

Add the values $8$ and $8$

$\int_{-2}^{2}-x^2dx+16$
6

Taking the constant out of the integral

$16-\int_{-2}^{2} x^2dx$
7

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\left[-\frac{x^{3}}{3}\right]_{-2}^{2}+16$
8

Evaluate the definite integral

$16-1\cdot \left(\frac{{\left(-2\right)}^{3}}{3}\right)\left(-1\right)+\frac{2^{3}}{3}\left(-1\right)$
9

Multiply $-1$ times $-1$

$16+\frac{{\left(-2\right)}^{3}}{3}\cdot 1+\frac{2^{3}}{3}\left(-1\right)$
10

Calculate the power

$16+\frac{-8}{3}\cdot 1+\frac{8}{3}\left(-1\right)$
11

Divide $8$ by $3$

$16-2.6667\cdot 1+2.6667\left(-1\right)$
12

Multiply $-1$ times $\frac{8}{3}$

$16-2.6667-2.6667$
13

Subtract the values $\frac{40}{3}$ and $-\frac{8}{3}$

$\frac{32}{3}$

Answer

$\frac{32}{3}$

Problem Analysis

Main topic:

Integral calculus

Time to solve it:

0.28 seconds

Views:

81