What is the integral from-1 to 1 of (2x^2-4x+3) ?

The integral from-1 to 1 of (2x^2-4x+3) is 22/3

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integral from-1 to 1 of (2x^2-4x+3)

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

\frac{22}{3}

Decimal

7.33333 \dots

Solution steps

\int_{- 1}^{1} 2 x^{2} - 4 x + 3 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_{- 1}^{1} 2 x^{2} d x - \int_{- 1}^{1} 4 x d x + \int_{- 1}^{1} 3 d x

\int_{- 1}^{1} 2 x^{2} d x = \frac{4}{3}

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

\int_{- 1}^{1} 3 d x = 6

= \frac{4}{3} - 0 + 6

Simplify

\frac{4}{3} - 0 + 6 : \frac{22}{3}

= \frac{22}{3}

\int_{2 y}^{2} cos (x^{2}) d x

\int_{0}^{3} (5 x + 2) d x

\int_{0}^{y} y^{2} d y

\int_{0}^{3} cos (x) d x

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

What is the integral from-1 to 1 of (2x^2-4x+3) ?
The integral from-1 to 1 of (2x^2-4x+3) is 22/3