What is the integral of 8e^{4y} ?

The integral of 8e^{4y} is 2e^{4y}+C

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integral of 8e^{4y}

\int 8 e^{4 y} d y

2 e^{4 y} + C

Solution steps

\int 8 e^{4 y} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int e^{4 y} d y

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= 8 \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 8 \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 y

= 8 \cdot \frac{1}{4} e^{4 y}

Simplify

8 \cdot \frac{1}{4} e^{4 y} : 2 e^{4 y}

= 2 e^{4 y}

Add a constant to the solution

= 2 e^{4 y} + C

\frac{1}{( s ^{2} + 3 )}

4 - x^{2}, x

- 2 y^{^{''}} + 2 y^{^{'}} = 0, y (1) = - 2, y^{^{'}} (- 2) = 3

x \to 0 lim (\frac{x}{3 + x - 3})

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