What is the integral from 0 to 4 of (2x^2+8) ?

The integral from 0 to 4 of (2x^2+8) is 224/3

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integral from 0 to 4 of (2x^2+8)

\int_{0}^{4} (2 x^{2} + 8) d x

\frac{224}{3}

Decimal

74.66666 \dots

Solution steps

\int_{0}^{4} 2 x^{2} + 8 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{4} 2 x^{2} d x + \int_{0}^{4} 8 d x

\int_{0}^{4} 2 x^{2} d x = \frac{128}{3}

\int_{0}^{4} 8 d x = 32

= \frac{128}{3} + 32

Simplify

\frac{128}{3} + 32 : \frac{224}{3}

= \frac{224}{3}

1 - e^{- x}

\frac{x ^{2}}{4 + x}

\frac{d}{d x} (3 cosh (2 x))

\int \frac{6 x ^{2} + 7 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

x \to 1 lim ([2 x])

What is the integral from 0 to 4 of (2x^2+8) ?
The integral from 0 to 4 of (2x^2+8) is 224/3