What is the integral from 0 to 1 of (5x^2-8x+4) ?

The integral from 0 to 1 of (5x^2-8x+4) is 5/3

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integral from 0 to 1 of (5x^2-8x+4)

\int_{0}^{1} (5 x^{2} - 8 x + 4) d x

\frac{5}{3}

Decimal

1.66666 \dots

Solution steps

\int_{0}^{1} 5 x^{2} - 8 x + 4 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}^{1} 5 x^{2} d x - \int_{0}^{1} 8 x d x + \int_{0}^{1} 4 d x

\int_{0}^{1} 5 x^{2} d x = \frac{5}{3}

\int_{0}^{1} 8 x d x = 4

\int_{0}^{1} 4 d x = 4

= \frac{5}{3} - 4 + 4

Simplify

= \frac{5}{3}

\int_{1}^{3} (x^{2} - 4 x + 8) d x

\int_{0}^{1} (- x) d x

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

\int_{0}^{4 π} sin^{2} (x) d x

\int_{0}^{\infty} \frac{3}{x ^{2}} d x

What is the integral from 0 to 1 of (5x^2-8x+4) ?
The integral from 0 to 1 of (5x^2-8x+4) is 5/3