What is the integral from 1 to x of (1-x) ?

The integral from 1 to x of (1-x) is x-1-(x^2-1)/2

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integral from 1 to x of (1-x)

\int_{1}^{x} (1 - x) d x

x - 1 - \frac{x ^{2} - 1}{2}

Solution steps

\int_{1}^{x} 1 - x 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}^{x} 1 d x - \int_{1}^{x} x d x

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

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

= x - 1 - \frac{x ^{2} - 1}{2}

\int_{2}^{7} \frac{2}{8 x - 11} d x

\int_{0}^{x} x cos (x + y) d y

\int_{\frac{1}{3}}^{3} 1 - \frac{1}{x ^{2}} d x

\int_{2}^{5} \frac{( x ^{4} + x ^{2} )}{3} d x

\int_{- 1}^{0} \frac{1}{1 + cos ( t )} d t

What is the integral from 1 to x of (1-x) ?
The integral from 1 to x of (1-x) is x-1-(x^2-1)/2