What is area y=-x^2+16,y=-1/4 x-1,4(y+2.3)=2.3x ?

The answer to area y=-x^2+16,y=-1/4 x-1,4(y+2.3)=2.3x is 97/384+16.60266…

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area y=-x^2+16,y=-1/4 x-1,4(y+2.3)=2.3x

area

y = - x^{2} + 16, y = - \frac{1}{4} x - 1, 4 (y + 2.3) = 2.3 x

\frac{97}{384} + 16.60266 \dots

Decimal

16.85527 \dots

Solution steps

Isolate

y

for

4 (y + 2.3) = 2.3 x : y = 0.575 x - 2.3

f_{1} (x) = - x^{2} + 16

f_{2} (x) = - \frac{1}{4} x - 1

To find the intersection points solve

f_{i} (x) = f_{j} (x)

- x^{2} + 16 = - \frac{1}{4} x - 1 : x = - 4, x = \frac{17}{4}

- x^{2} + 16 = 0.575 x - 2.3 : x = - \frac{183}{40}, x = 4

- \frac{1}{4} x - 1 = 0.575 x - 2.3 : x = \frac{52}{33}

= \int_{- \frac{183}{40}}^{- 4} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 4}^{\frac{52}{33}} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{\frac{52}{33}}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{4}^{\frac{17}{4}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- \frac{183}{40}}^{- 4} - x^{2} + 16 - (0.575 x - 2.3) d x + \int_{- 4}^{\frac{52}{33}} 0.575 x - 2.3 - (- \frac{1}{4} x - 1) d x + \int_{\frac{52}{33}}^{4} - \frac{1}{4} x - 1 - (0.575 x - 2.3) d x + \int_{4}^{\frac{17}{4}} - x^{2} + 16 - (- \frac{1}{4} x - 1) d x

\int_{- \frac{183}{40}}^{- 4} - x^{2} + 16 - (0.575 x - 2.3) d x = 1.35418 \dots

\int_{- 4}^{\frac{52}{33}} 0.575 x - 2.3 - (- \frac{1}{4} x - 1) d x = 12.82424 \dots

\int_{\frac{52}{33}}^{4} - \frac{1}{4} x - 1 - (0.575 x - 2.3) d x = 2.42424 \dots

\int_{4}^{\frac{17}{4}} - x^{2} + 16 - (- \frac{1}{4} x - 1) d x = \frac{97}{384}

= 1.35418 \dots + 12.82424 \dots + 2.42424 \dots + \frac{97}{384}

Simplify

= \frac{97}{384} + 16.60266 \dots

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x \to 7 lim (\frac{7 - x}{7 - x})

x \to 6 lim (- 110)

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\frac{d}{d x} (\frac{4}{3} π \cdot x^{3})

What is area y=-x^2+16,y=-1/4 x-1,4(y+2.3)=2.3x ?
The answer to area y=-x^2+16,y=-1/4 x-1,4(y+2.3)=2.3x is 97/384+16.60266…