What is the integral from 2 to 5 of (6x^2-3x+5) ?

The integral from 2 to 5 of (6x^2-3x+5) is 435/2

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integral from 2 to 5 of (6x^2-3x+5)

\int_{2}^{5} (6 x^{2} - 3 x + 5) d x

\frac{435}{2}

Decimal

217.5

Solution steps

\int_{2}^{5} 6 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_{2}^{5} 6 x^{2} d x - \int_{2}^{5} 3 x d x + \int_{2}^{5} 5 d x

\int_{2}^{5} 6 x^{2} d x = 234

\int_{2}^{5} 3 x d x = \frac{63}{2}

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

= 234 - \frac{63}{2} + 15

Simplify

234 - \frac{63}{2} + 15 : \frac{435}{2}

= \frac{435}{2}

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

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

\int_{0}^{2} 2 t 4 t^{2} + 1 d t

\int_{- 1}^{x} \frac{d}{1} (x^{2} - x) d x

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

What is the integral from 2 to 5 of (6x^2-3x+5) ?
The integral from 2 to 5 of (6x^2-3x+5) is 435/2