What is area y=1+sqrt(x),y=((3+x))/3 ?

The answer to area y=1+sqrt(x),y=((3+x))/3 is 9/2

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area y=1+sqrt(x),y=((3+x))/3

area

y = 1 + x, y = \frac{( 3 + x )}{3}

\frac{9}{2}

Decimal

4.5

Solution steps

Apply the area formula:

\int_{0}^{9} 1 + x - \frac{1}{3} (3 + x) d x

Solve

\int_{0}^{9} 1 + x - \frac{1}{3} (3 + x) d x : \frac{9}{2}

The area is:

= \frac{9}{2}

\int 6 - x d x

\frac{s - 3}{s ^{2} - 4 s + 4}

\frac{d}{d x} (ln (\frac{x ^{2}}{3 x + 1}))

\int (2 + sin^{2} (θ))^{2} cos (θ) d θ

y^{^{'}} - \frac{4}{x} y = 3 x - 2

What is area y=1+sqrt(x),y=((3+x))/3 ?
The answer to area y=1+sqrt(x),y=((3+x))/3 is 9/2