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Popular >

integral from 0 to 1 of (-x+1)

Solution

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

Solution

\frac{1}{2}

+1

Decimal Notation

0.5

Solution steps

\int_{0}^{1} - x + 1 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} 1 d x

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

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

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

Simplify

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

= \frac{1}{2}

Popular Examples

integral from 0 to 5 of t^2-t

\int_{0}^{5} t^{2} - t d t

integral from 0 to 2 of \sqrt[3]{x}

integral from 4.6 to 7 of (12x^2-6x+4)

\int_{4.6}^{7} (12 x^{2} - 6 x + 4) d x

integral from 0 to 1 of 2(1-y)

\int_{0}^{1} 2 (1 - y) d y

integral from 0 to x of (t^2)/(1+t^3)

\int_{0}^{x} \frac{t ^{2}}{1 + t ^{3}} d t

Frequently Asked Questions (FAQ)

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