What is the integral from 1 to 3 of (2x^2-3x+5) ?

The integral from 1 to 3 of (2x^2-3x+5) is 46/3

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integral from 1 to 3 of (2x^2-3x+5)

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

\frac{46}{3}

Solution steps

\int_{1}^{3} 2 x^{2} - 3 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}^{3} 2 x^{2} d x - \int_{1}^{3} 3 x d x + \int_{1}^{3} 5 d x

\int_{1}^{3} 2 x^{2} d x = \frac{52}{3}

\int_{1}^{3} 3 x d x = 12

\int_{1}^{3} 5 d x = 10

= \frac{52}{3} - 12 + 10

Simplify

\frac{52}{3} - 12 + 10 : \frac{46}{3}

= \frac{46}{3}

e^{13 t} sin (t)

x \frac{d y}{d x} + y = 6 x + 1, y (1) = 9

\frac{d}{d x} (4 sec (x) - 5 x)

x \to 0 + lim (tan^{x} (7 x))

5 e^{- t} sin (2 t)

What is the integral from 1 to 3 of (2x^2-3x+5) ?
The integral from 1 to 3 of (2x^2-3x+5) is 46/3