What is area y=sqrt(9-x^2),0<= x<= 2 ?

The answer to area y=sqrt(9-x^2),0<= x<= 2 is 9/2 arcsin(2/3)+sqrt(5)

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area y=sqrt(9-x^2),0<= x<= 2

area

y = 9 - x^{2}, 0 \leq x \leq 2

\frac{9}{2} arcsin (\frac{2}{3}) + 5

Decimal Notation

5.51984 \dots

Solution steps

Apply the area formula:

\int_{0}^{2} 9 - x^{2} d x

Solve

\int_{0}^{2} 9 - x^{2} d x : \frac{9}{2} arcsin (\frac{2}{3}) + 5

The area is:

= \frac{9}{2} arcsin (\frac{2}{3}) + 5

f (x) = 7 e^{x^{2} - 9 x + 8}, at x = 0

y = 164 4 x^{4} + 4

y^{^{''}} + 6 y^{^{'}} + 5 y = 10

\int \frac{1}{h} d h

\frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 5

What is area y=sqrt(9-x^2),0<= x<= 2 ?
The answer to area y=sqrt(9-x^2),0<= x<= 2 is 9/2 arcsin(2/3)+sqrt(5)