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

The integral from-1 to 2 of (2x^2-4x+5) is 15

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

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

15

Solution steps

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

\int_{- 1}^{2} 2 x^{2} d x = 6

\int_{- 1}^{2} 4 x d x = 6

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

= 6 - 6 + 15

Simplify

= 15

\frac{\partial}{\partial x} (2 x^{4} y - 2)

\int \frac{1}{1 + e ^{6 x}} d x

n = 1 \sum \infty \frac{5}{n + 2}

\int (x^{3} + 3) d x

e^{x} \frac{d y}{d x} = e^{- y} + e^{2 x - y}

What is the integral from-1 to 2 of (2x^2-4x+5) ?
The integral from-1 to 2 of (2x^2-4x+5) is 15