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Popular Calculus >

integral from 2 to 1 of (1+z/2)

Solution

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

Solution

- \frac{7}{4}

+1

Decimal

- 1.75

Solution steps

\int_{2}^{1} 1 + \frac{z}{2} d z

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} 1 + \frac{z}{2} d z

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} 1 d z + \int_{1}^{2} \frac{z}{2} d z)

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

\int_{1}^{2} \frac{z}{2} d z = \frac{3}{4}

= - (1 + \frac{3}{4})

Join

1 + \frac{3}{4} : \frac{7}{4}

= - \frac{7}{4}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 2 to 1 of (1+z/2) ?
The integral from 2 to 1 of (1+z/2) is -7/4