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

The answer to area y= 1/3 x^3+2,y=x+1,0,2 is 4/3

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

area

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{2} \frac{1}{3} x^{3} + 2 - (x + 1) d x : \frac{4}{3}

The area is:

= \frac{4}{3}

\int \frac{sin ^{3} ( x )}{( x )} d x

\int_{- 1}^{1} 1 0^{x} d x

(\frac{3}{x})^{^{'}}

\frac{d}{d x} (5 x^{2} ln (x))

(x^{3} - 16 x)^{14}

What is area y= 1/3 x^3+2,y=x+1,0,2 ?
The answer to area y= 1/3 x^3+2,y=x+1,0,2 is 4/3