What is area x= 1/3 (y^2+2)^{3/2},1<= y<= 5 ?

The answer to area x= 1/3 (y^2+2)^{3/2},1<= y<= 5 is (ln((26+15sqrt(3))/(2+3^{1/2)})+148sqrt(3))/4

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area x= 1/3 (y^2+2)^{3/2},1<= y<= 5

area

x = \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}, 1 \leq y \leq 5

\frac{ln ( \frac{26 + 15 3}{2 + 3 ^{\frac{1}{2}}} ) + 148 3}{4}

Decimal Notation

64.74435 \dots

Solution steps

Apply the area formula:

\int_{1}^{5} \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}} d y

Solve

\int_{1}^{5} \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}} d y : \frac{ln ( \frac{26 + 15 3}{2 + 3 ^{\frac{1}{2}}} ) + 148 3}{4}

The area is:

= \frac{ln ( \frac{26 + 15 3}{2 + 3 ^{\frac{1}{2}}} ) + 148 3}{4}

f^{^{'}} (x) = 12 x^{2} - 6 x + 1, f (1) = 5

\int \frac{5}{x ( 1 + x )} d x

\frac{1}{2}

x^{2} 1 + x^{2}

\int (- x^{2} - 3 x + 3)^{2} d x

What is area x= 1/3 (y^2+2)^{3/2},1<= y<= 5 ?
The answer to area x= 1/3 (y^2+2)^{3/2},1<= y<= 5 is (ln((26+15sqrt(3))/(2+3^{1/2)})+148sqrt(3))/4