What is the integral from 0 to 2 of 2pix(1+(x^2)/2) ?

The integral from 0 to 2 of 2pix(1+(x^2)/2) is 8pi

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integral from 0 to 2 of 2pix(1+(x^2)/2)

\int_{0}^{2} 2 π x (1 + \frac{x ^{2}}{2}) d x

8 π

Decimal Notation

25.13274 \dots

Solution steps

\int_{0}^{2} 2 π x (1 + \frac{x ^{2}}{2}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int_{0}^{2} x (1 + \frac{x ^{2}}{2}) d x

Apply u-substitution

: \int_{1}^{3} u d u

= 2 π \cdot \int_{1}^{3} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{1}^{3}

= 2 π [\frac{u ^{2}}{2}]_{1}^{3}

Compute the boundaries:

4

= 2 π 4

Simplify

= 8 π

x \to 1 lim (x^{2} + 8 x - 2)

\frac{d}{d x} (2 x^{2} cos (2 x))

y = 2 cos (x), at x = \frac{π}{3}

y^{2} + 5 x = x^{2} y - 5, (3, 5)

x = y^{2}, x = 3 - 2 y^{2}

What is the integral from 0 to 2 of 2pix(1+(x^2)/2) ?
The integral from 0 to 2 of 2pix(1+(x^2)/2) is 8pi