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Popular >

integral from 1 to 3 of (4z-z^2)

Solution

\int_{1}^{3} (4 z - z^{2}) d z

Solution

\frac{22}{3}

+1

Decimal Notation

7.33333 \dots

Solution steps

\int_{1}^{3} 4 z - z^{2} d z

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}^{3} 4 z d z - \int_{1}^{3} z^{2} d z

\int_{1}^{3} 4 z d z = 16

\int_{1}^{3} z^{2} d z = \frac{26}{3}

= 16 - \frac{26}{3}

Simplify

16 - \frac{26}{3} : \frac{22}{3}

= \frac{22}{3}

Popular Examples

y^{^{'}} + 4 y = 2 x

integral of (x^2)/((169+x^2)^2)

\int \frac{x ^{2}}{( 169 + x ^{2} ) ^{2}} d x

sum from n=0 to infinity of nz^n

n = 0 \sum \infty n z^{n}

integral of ((2x+1))/((x+1)(x-2)(x-3))

\int \frac{( 2 x + 1 )}{( x + 1 ) ( x - 2 ) ( x - 3 )} d x

integral of sqrt(7-2x)

\int 7 - 2 x d x

Frequently Asked Questions (FAQ)

What is the integral from 1 to 3 of (4z-z^2) ?
The integral from 1 to 3 of (4z-z^2) is 22/3