What is area y=e^{3x},y=e^{4x},x=1 ?

The answer to area y=e^{3x},y=e^{4x},x=1 is (3e^4-4e^3+1)/(12)

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area y=e^{3x},y=e^{4x},x=1

area

y = e^{3 x}, y = e^{4 x}, x = 1

\frac{3 e ^{4} - 4 e ^{3} + 1}{12}

Decimal Notation

7.03769 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} e^{3 x} - e^{4 x} d x

Solve

\int_{0}^{1} e^{3 x} - e^{4 x} d x : \frac{3 e ^{4} - 4 e ^{3} + 1}{12}

The area is:

= \frac{3 e ^{4} - 4 e ^{3} + 1}{12}

x = 7 - 7 y^{2}, x = 7 y^{2} - 7

2 t y^{2} d t + 2 t^{2} y d y = 0

\int (e^{c o s (x)} sin (x)) d x

x \to 0 lim (ln (x^{2}) 2 e^{x} + 1)

\frac{d}{d x} (cos^{s i n (x)} (x))

What is area y=e^{3x},y=e^{4x},x=1 ?
The answer to area y=e^{3x},y=e^{4x},x=1 is (3e^4-4e^3+1)/(12)