What is area y=2^x,y=4^x,y=4 ?

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

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area y=2^x,y=4^x,y=4

area

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

- \frac{3}{2 ln ( 2 )} + 4

Decimal Notation

1.83595 \dots

Solution steps

To find the intersection points solve

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

2^{x} = 4^{x} : x = 0

2^{x} = 4 : x = 2

4^{x} = 4 : x = 1

= \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{1} ∣ 2^{x} - 4^{x} ∣ d x + \int_{1}^{2} ∣ 2^{x} - 4 ∣ d x

\int_{0}^{1} ∣ 2^{x} - 4^{x} ∣ d x = \frac{1}{2 ln ( 2 )}

\int_{1}^{2} ∣ 2^{x} - 4 ∣ d x = - \frac{2}{ln ( 2 )} + 4

= \frac{1}{2 ln ( 2 )} - \frac{2}{ln ( 2 )} + 4

Simplify

= - \frac{3}{2 ln ( 2 )} + 4

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