What is area y= 6/(4-3x)-1,(-4,1) ?

The answer to area y= 6/(4-3x)-1,(-4,1) is ln(81/16)+5/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y= 6/(4-3x)-1,(-4,1)

area

y = \frac{6}{4 - 3 x} - 1, (- 4, 1)

ln (\frac{81}{16}) + \frac{5}{3}

Decimal

3.28852 \dots

Solution steps

Apply the area formula:

\int_{- 4}^{1} \frac{6}{4 - 3 x} - 1 d x

Solve

\int_{- 4}^{1} \frac{6}{4 - 3 x} - 1 d x : ln (\frac{81}{16}) + \frac{5}{3}

The area is:

= ln (\frac{81}{16}) + \frac{5}{3}

\frac{\partial}{\partial x} (2 x^{2} y + cos (x^{4}))

\int \frac{1}{10 - x} d x

\frac{\partial}{\partial z} (x tan (y) - z e^{z})

\frac{d y}{d t} + 3 y = t + e^{- 2 t}

\int \frac{4}{3 2 x + 1} d x

What is area y= 6/(4-3x)-1,(-4,1) ?
The answer to area y= 6/(4-3x)-1,(-4,1) is ln(81/16)+5/3