What is area y=x^2-4x+3,y=3+4x-x^2 ?

The answer to area y=x^2-4x+3,y=3+4x-x^2 is 64/3

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area y=x^2-4x+3,y=3+4x-x^2

area

y = x^{2} - 4 x + 3, y = 3 + 4 x - x^{2}

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} x^{2} - 4 x + 3 - (3 + 4 x - x^{2}) d x

Solve

\int_{0}^{4} x^{2} - 4 x + 3 - (3 + 4 x - x^{2}) d x : \frac{64}{3}

The area is:

= \frac{64}{3}

x \to - \infty lim (e^{x})

cos (\frac{π x}{4})

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

x = 3 y, x = y^{2} - 4

\int x^{2} e^{- λ x} d x

What is area y=x^2-4x+3,y=3+4x-x^2 ?
The answer to area y=x^2-4x+3,y=3+4x-x^2 is 64/3