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Popular >

integral from 1 to 4 of (x-1)

Solution

\int_{1}^{4} (x - 1) d x

Solution

\frac{9}{2}

+1

Solution steps

\int_{1}^{4} 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_{1}^{4} x d x - \int_{1}^{4} 1 d x

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 1 d x = 3

= \frac{15}{2} - 3

Simplify

\frac{15}{2} - 3 : \frac{9}{2}

= \frac{9}{2}

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\int_{0}^{1} \frac{6}{1 + x ^{2}} d x

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f (x) = 3 cot^{4} (x), at x = \frac{π}{4}

integral of 3/(x^2sqrt(x^2+9))

\int \frac{3}{x ^{2} x ^{2} + 9} d x

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\frac{d}{d x} (x^{e^{2}})

Frequently Asked Questions (FAQ)

What is the integral from 1 to 4 of (x-1) ?
The integral from 1 to 4 of (x-1) is 9/2