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Popular Calculus >

integral from 0 to 1 of (x+1/2)

Solution

\int_{0}^{1} (x + \frac{1}{2}) d x

Solution

1

Solution steps

\int_{0}^{1} x + \frac{1}{2} 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_{0}^{1} x d x + \int_{0}^{1} \frac{1}{2} d x

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} \frac{1}{2} d x = \frac{1}{2}

= \frac{1}{2} + \frac{1}{2}

Simplify

\frac{1}{2} + \frac{1}{2} : 1

= 1

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of (x+1/2) ?
The integral from 0 to 1 of (x+1/2) is 1