What is area y=(x-2)^2-4,y=x+1 ?

The answer to area y=(x-2)^2-4,y=x+1 is sqrt(29)+(23sqrt(29))/6

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area y=(x-2)^2-4,y=x+1

area

y = (x - 2)^{2} - 4, y = x + 1

29 + \frac{23 29}{6}

Decimal Notation

26.02829 \dots

Solution steps

Apply the area formula:

\int_{\frac{5 - 29}{2}}^{\frac{5 + 29}{2}} (x - 2)^{2} - 4 - (x + 1) d x

Solve

\int_{\frac{5 - 29}{2}}^{\frac{5 + 29}{2}} (x - 2)^{2} - 4 - (x + 1) d x : 29 + \frac{23 29}{6}

The area is:

= 29 + \frac{23 29}{6}

\int \frac{cos ( \frac{π}{x} )}{x ^{2}} d x

\frac{d}{d x} (x^{2} - 16 x)

- 7 e^{(x + 2)} + e^{5}

(6, - 7), (8, 5)

\frac{\partial}{\partial y} (4 y^{2} - 20)

What is area y=(x-2)^2-4,y=x+1 ?
The answer to area y=(x-2)^2-4,y=x+1 is sqrt(29)+(23sqrt(29))/6