What is area y= 3/x+2x^3,2<= x<= 3 ?

The answer to area y= 3/x+2x^3,2<= x<= 3 is 3(ln(3)-ln(2))+65/2

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area y= 3/x+2x^3,2<= x<= 3

area

y = \frac{3}{x} + 2 x^{3}, 2 \leq x \leq 3

3 (ln (3) - ln (2)) + \frac{65}{2}

Decimal Notation

33.71639 \dots

Solution steps

Apply the area formula:

\int_{2}^{3} \frac{3}{x} + 2 x^{3} d x

Solve

\int_{2}^{3} \frac{3}{x} + 2 x^{3} d x : 3 (ln (3) - ln (2)) + \frac{65}{2}

The area is:

= 3 (ln (3) - ln (2)) + \frac{65}{2}

f (x) = - 2 x^{3}, (1, - 2)

\frac{d}{d x} (x (x + 12))

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

f (x) = \frac{1}{6 x ^{3}}

f (x) = \frac{2}{x - 6}, at x = 8

What is area y= 3/x+2x^3,2<= x<= 3 ?
The answer to area y= 3/x+2x^3,2<= x<= 3 is 3(ln(3)-ln(2))+65/2