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

The integral from 2 to 4 of (2x^2-3x+1) is 64/3

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

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

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

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

\int_{2}^{4} 2 x^{2} d x = \frac{112}{3}

\int_{2}^{4} 3 x d x = 18

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

= \frac{112}{3} - 18 + 2

Simplify

\frac{112}{3} - 18 + 2 : \frac{64}{3}

= \frac{64}{3}

x \to - 2 lim (\frac{3}{x + 2})

\int 1 - u^{2} d u

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = \frac{82}{5}

\int 1 - 81 x^{2} d x

\int 7 cos^{2} (x) d x

What is the integral from 2 to 4 of (2x^2-3x+1) ?
The integral from 2 to 4 of (2x^2-3x+1) is 64/3