What is the integral from 2 to 3.5 of sqrt(7-2x) ?

The integral from 2 to 3.5 of sqrt(7-2x) is square root of 3

Popular >

integral from 2 to 3.5 of sqrt(7-2x)

\int_{2}^{3.5} 7 - 2 x d x

3

Decimal Notation

1.73205 \dots

Solution steps

\int_{2}^{3.5} 7 - 2 x d x

Apply u-substitution

: \int_{3}^{0} - \frac{u}{2} d u

= \int_{3}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{3} - \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{0}^{3} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{3}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{3})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{3}

Compute the boundaries:

23

= \frac{1}{2} \cdot 23

Simplify

= 3

\int_{- 1}^{2} 3 x^{2} e^{x^{3} + 5} d x

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

\int_{0}^{1} \frac{1}{9 - x ^{2}} d x

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

\int_{0}^{1} x^{4} \cdot e^{- x} d x

What is the integral from 2 to 3.5 of sqrt(7-2x) ?
The integral from 2 to 3.5 of sqrt(7-2x) is square root of 3