What is the integral from 1 to 4 of (-3)/((2p+1)^2) ?

The integral from 1 to 4 of (-3)/((2p+1)^2) is -1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of (-3)/((2p+1)^2)

\int_{1}^{4} \frac{- 3}{( 2 p + 1 ) ^{2}} d p

- \frac{1}{3}

Decimal Notation

- 0.33333 \dots

Solution steps

\int_{1}^{4} \frac{- 3}{( 2 p + 1 ) ^{2}} d p

Take the constant out:

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

= - 3 \cdot \int_{1}^{4} \frac{1}{( 2 p + 1 ) ^{2}} d p

Apply u-substitution

: \int_{3}^{9} \frac{1}{2 u ^{2}} d u

= - 3 \cdot \int_{3}^{9} \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= - 3 \cdot \frac{1}{2} \cdot \int_{3}^{9} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{9}

= - 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{9}

Simplify

- 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{9} : - \frac{3}{2} [- \frac{1}{u}]_{3}^{9}

= - \frac{3}{2} [- \frac{1}{u}]_{3}^{9}

Compute the boundaries:

\frac{2}{9}

= - \frac{3}{2} \cdot \frac{2}{9}

Simplify

= - \frac{1}{3}

\int_{0}^{8} 3 x^{- \frac{1}{3}} d x

\int_{- 2}^{0} x cos (\frac{nπ x}{2}) d x

\int_{2}^{3} (7 x^{2} + 5 x + 4) d x

\int_{0}^{1} x^{7} \cdot e^{- x} d x

\int_{1}^{4} (- 2 x^{2} + 12 x - 16) d x

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