What is the integral from 0 to 1 of (7x^2-8x+6) ?

The integral from 0 to 1 of (7x^2-8x+6) is 13/3

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integral from 0 to 1 of (7x^2-8x+6)

\int_{0}^{1} (7 x^{2} - 8 x + 6) d x

\frac{13}{3}

Decimal

4.33333 \dots

Solution steps

\int_{0}^{1} 7 x^{2} - 8 x + 6 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} 7 x^{2} d x - \int_{0}^{1} 8 x d x + \int_{0}^{1} 6 d x

\int_{0}^{1} 7 x^{2} d x = \frac{7}{3}

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

\int_{0}^{1} 6 d x = 6

= \frac{7}{3} - 4 + 6

Simplify

\frac{7}{3} - 4 + 6 : \frac{13}{3}

= \frac{13}{3}

\int_{1}^{5} x^{2} - 2 x - 15 d x

\int_{1}^{5} (3 x^{2}) d x

\int_{3}^{6} (x - 3)^{2} d x

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

\int_{0}^{4} \frac{3}{5 x + 7} d x

What is the integral from 0 to 1 of (7x^2-8x+6) ?
The integral from 0 to 1 of (7x^2-8x+6) is 13/3