What is the integral from 1 to 2 of x^2+6/2 x ?

The integral from 1 to 2 of x^2+6/2 x is 41/6

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integral from 1 to 2 of x^2+6/2 x

\int_{1}^{2} x^{2} + \frac{6}{2} x d x

\frac{41}{6}

Decimal Notation

6.83333 \dots

Solution steps

\int_{1}^{2} x^{2} + \frac{6}{2} x d x

\frac{6}{2} x = 3 x

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} 3 x d x = \frac{9}{2}

= \frac{7}{3} + \frac{9}{2}

Simplify

\frac{7}{3} + \frac{9}{2} : \frac{41}{6}

= \frac{41}{6}

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

\int_{0}^{4} 15 - 2 x - 7 d x

What is the integral from 1 to 2 of x^2+6/2 x ?
The integral from 1 to 2 of x^2+6/2 x is 41/6