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Popular >

integral from 0 to 2 of 1-c|1-x|

Solution

\int_{0}^{2} 1 - c ∣ 1 - x ∣ d x

Solution

- c + 2

Solution steps

\int_{0}^{2} 1 - c ∣ 1 - x ∣ d x

Eliminate Absolutes

= \int_{0}^{1} 1 - c (1 - x) d x + \int_{1}^{2} 1 - c (- 1 + x) d x

\int_{0}^{1} 1 - c (1 - x) d x = 1 - \frac{1}{2} c

\int_{1}^{2} 1 - c (- 1 + x) d x = 1 - \frac{1}{2} c

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

Simplify

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

= - c + 2

Popular Examples

integral from-1 to 1 of-xsin(npix)

\int_{- 1}^{1} - x sin (nπ x) d x

integral from 2 to 5 of 3/(sqrt(x))

\int_{2}^{5} \frac{3}{x} d x

integral from 0 to 1 of (e^{2x}-5)

\int_{0}^{1} (e^{2 x} - 5) d x

integral from 0 to y of 2e^{-2x}

\int_{0}^{y} 2 e^{- 2 x} d x

integral from 1 to infinity of 1/(2x+6)

\int_{1}^{\infty} \frac{1}{2 x + 6} d x

Frequently Asked Questions (FAQ)

What is the integral from 0 to 2 of 1-c|1-x| ?
The integral from 0 to 2 of 1-c|1-x| is -c+2