What is the integral from a to b of 1/(b-a) ?

The integral from a to b of 1/(b-a) is 1

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integral from a to b of 1/(b-a)

\int_{a}^{b} \frac{1}{b - a}

1

Solution steps

\int_{a}^{b} \frac{1}{b - a} d x

Integral of a constant:

\int a d x = a x

= [\frac{1}{b - a} x]_{a}^{b}

Multiply

\frac{1}{b - a} x : \frac{x}{b - a}

= [\frac{x}{b - a}]_{a}^{b}

Compute the boundaries:

\frac{b}{b - a} - \frac{a}{b - a}

= \frac{b}{b - a} - \frac{a}{b - a}

Simplify

= 1

16 y^{^{''}} + 24 y^{^{'}} + 9 y = 0, y (2) = 4, y^{^{'}} (- 2) = - 3

- 3 cos (- 3 x)

\frac{d}{d x} (x - 1 - ln (x))

x^{2} (1 - y) \frac{d y}{d x} + y^{2} (1 + x) = 0

\frac{d y}{d t} - 2 y = 8 e^{4 t}

What is the integral from a to b of 1/(b-a) ?
The integral from a to b of 1/(b-a) is 1