What is the integral from 0 to 1 of (5x^2-4x+7) ?

The integral from 0 to 1 of (5x^2-4x+7) is 20/3

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integral from 0 to 1 of (5x^2-4x+7)

\int_{0}^{1} (5 x^{2} - 4 x + 7) d x

\frac{20}{3}

Decimal

6.66666 \dots

Solution steps

\int_{0}^{1} 5 x^{2} - 4 x + 7 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} 4 x d x + \int_{0}^{1} 7 d x

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

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} 7 d x = 7

= \frac{5}{3} - 2 + 7

Simplify

\frac{5}{3} - 2 + 7 : \frac{20}{3}

= \frac{20}{3}

\int_{5}^{1} x^{2} x - 1 d x

\int_{- 6}^{2} x - 12 - 4 x d x

\int_{0.5}^{0.75} 12 x^{2} d x

\int_{1}^{3} \frac{3}{7 x - 2} d x

\int_{x}^{4} \frac{6 ( y - x )}{64} d y

What is the integral from 0 to 1 of (5x^2-4x+7) ?
The integral from 0 to 1 of (5x^2-4x+7) is 20/3