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Popular >

integral from 0 to 1 of (t-1)

Solution

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

Solution

- \frac{1}{2}

+1

Decimal Notation

- 0.5

Solution steps

\int_{0}^{1} t - 1 d t

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} t d t - \int_{0}^{1} 1 d t

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

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

= \frac{1}{2} - 1

Simplify

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

= - \frac{1}{2}

Popular Examples

integral from 0 to 1 of x^{1/4}-x^{1/2}

\int_{0}^{1} x^{\frac{1}{4}} - x^{\frac{1}{2}} d x

integral from 0 to 3 of pi(3x-x^2)^2

\int_{0}^{3} π (3 x - x^{2})^{2} d x

integral from 4 to 6 of 1/(sqrt(x^2-9))

\int_{4}^{6} \frac{1}{x ^{2} - 9} d x

integral from 1 to infinity of 1/(x^{0.99)}

\int_{1}^{\infty} \frac{1}{x ^{0.99}} d x

integral from 0 to 4 of (6+7sqrt(x))

\int_{0}^{4} (6 + 7 x) d x

Frequently Asked Questions (FAQ)

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