What is the integral from 0 to h of b/h (h-y) ?

The integral from 0 to h of b/h (h-y) is (bh)/2

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integral from 0 to h of b/h (h-y)

\int_{0}^{h} \frac{b}{h} (h - y) d y

\frac{bh}{2}

Solution steps

\int_{0}^{h} \frac{b}{h} (h - y) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{b}{h} \cdot \int_{0}^{h} h - y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{b}{h} (\int_{0}^{h} h d y - \int_{0}^{h} y d y)

\int_{0}^{h} h d y = h^{2}

\int_{0}^{h} y d y = \frac{h ^{2}}{2}

= \frac{b}{h} (h^{2} - \frac{h ^{2}}{2})

Simplify

\frac{b}{h} (h^{2} - \frac{h ^{2}}{2}) : \frac{bh}{2}

= \frac{bh}{2}

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

\int_{0}^{π} cos (\frac{1}{2}) x d x

x \to \frac{π}{2} lim (\frac{cos ( x )}{1 - sin ( x )})

x \to \frac{π}{2} lim (\frac{ln ( sin ( x ) )}{( π - 2 x ) ^{2}})

x \to 3 lim ((2 x - 4)^{4} (x - 2))

What is the integral from 0 to h of b/h (h-y) ?
The integral from 0 to h of b/h (h-y) is (bh)/2