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

The integral from 1 to 2 of ((x^2)/4+1) is 19/12

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

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

\frac{19}{12}

Decimal

1.58333 \dots

Solution steps

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

\int_{1}^{2} \frac{x ^{2}}{4} d x = \frac{7}{12}

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

= \frac{7}{12} + 1

Simplify

\frac{7}{12} + 1 : \frac{19}{12}

= \frac{19}{12}

What is the integral from 1 to 2 of ((x^2)/4+1) ?
The integral from 1 to 2 of ((x^2)/4+1) is 19/12