What is area y=-x+4,y=x^2,x=0,x=4 ?

The answer to area y=-x+4,y=x^2,x=0,x=4 is 2(sqrt(17)-1)+(67+5sqrt(17))/6

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area y=-x+4,y=x^2,x=0,x=4

area

y = - x + 4, y = x^{2}, x = 0, x = 4

2 (17 - 1) + \frac{67 + 5 17}{6}

Decimal Notation

20.84879 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{4} - x + 4 - x^{2} d x : 2 (17 - 1) + \frac{67 + 5 17}{6}

The area is:

= 2 (17 - 1) + \frac{67 + 5 17}{6}

\int 3 x^{- 4} (2 + 4 x^{- 3})^{- 7} d x

f^{^{'}} (t) - f (t) = - 4 t

\int_{- 8}^{1} \frac{1}{x ^{\frac{1}{3}}} d x

y^{^{'}} + 8 (tan (8 x)) y = - 2 cos (8 x)

\frac{( t ^{6} )}{( t ^{3} + 1 ) ^{2}}

What is area y=-x+4,y=x^2,x=0,x=4 ?
The answer to area y=-x+4,y=x^2,x=0,x=4 is 2(sqrt(17)-1)+(67+5sqrt(17))/6