What is area y= 2/3 x^{3/2}+3,[0,2] ?

The answer to area y= 2/3 x^{3/2}+3,[0,2] is (16*2^{1/2})/(15)+6

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area y= 2/3 x^{3/2}+3,[0,2]

area

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

\frac{16 \cdot 2 ^{\frac{1}{2}}}{15} + 6

Decimal Notation

7.50849 \dots

Solution steps

Apply the area formula:

\int_{0}^{2} \frac{2}{3} x^{\frac{3}{2}} + 3 d x

Solve

\int_{0}^{2} \frac{2}{3} x^{\frac{3}{2}} + 3 d x : \frac{16 \cdot 2 ^{\frac{1}{2}}}{15} + 6

The area is:

= \frac{16 \cdot 2 ^{\frac{1}{2}}}{15} + 6

y = x^{n}

\frac{6}{y} d x = (e^{3 x} + 1) d y

(x + 1) (\frac{d y}{d x}) = x + 6

f (x) = 1 + x^{2}

\frac{d}{d x} (\frac{- 7}{x ^{2}})

What is area y= 2/3 x^{3/2}+3,[0,2] ?
The answer to area y= 2/3 x^{3/2}+3,[0,2] is (16*2^{1/2})/(15)+6