What is area y= 1/2 x^3+2,y=x+1,x=0,x=2 ?

The answer to area y= 1/2 x^3+2,y=x+1,x=0,x=2 is 2

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area y= 1/2 x^3+2,y=x+1,x=0,x=2

area

y = \frac{1}{2} x^{3} + 2, y = x + 1, x = 0, x = 2

2

Solution steps

Apply the area formula:

\int_{0}^{2} \frac{1}{2} x^{3} + 2 - (x + 1) d x

Solve

\int_{0}^{2} \frac{1}{2} x^{3} + 2 - (x + 1) d x : 2

The area is:

= 2

x \to \infty lim (1 + \frac{2 ^{x + 1}}{2 ^{x} - 1})

\int \frac{3}{y 4 - ( ln ( y ) ) ^{2}} d y

n = 1 \sum \infty \frac{1}{( n + 7 ) n}

f (x) = x - 2, (6, 2)

\frac{d y}{d x} 7 + e^{- 5 x} y = 0

What is area y= 1/2 x^3+2,y=x+1,x=0,x=2 ?
The answer to area y= 1/2 x^3+2,y=x+1,x=0,x=2 is 2