What is the integral of 7e^{3x} ?

The integral of 7e^{3x} is 7/3 e^{3x}+C

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integral of 7e^{3x}

\int 7 e^{3 x} d x

\frac{7}{3} e^{3 x} + C

Solution steps

\int 7 e^{3 x} d x

Take the constant out:

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

= 7 \cdot \int e^{3 x} d x

Apply u-substitution

= 7 \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 7 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x

= 7 \cdot \frac{1}{3} e^{3 x}

Simplify

7 \cdot \frac{1}{3} e^{3 x} : \frac{7}{3} e^{3 x}

= \frac{7}{3} e^{3 x}

Add a constant to the solution

= \frac{7}{3} e^{3 x} + C

x - 1, y = 0, x = 6

f (t) = 4 e^{\frac{t}{2}}

f (x) = x^{2} - 5, at x = 3

\frac{\partial}{\partial y} (\frac{2}{2 x + 3 y})

f (x) = x^{2} sin (x)