What is the integral of e^{0.02x} ?

The integral of e^{0.02x} is 50e^{0.02x}+C

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integral of e^{0.02x}

\int e^{0.02 x} d x

50 e^{0.02 x} + C

Solution steps

\int e^{0.02 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{0.02} d u

= \int e^{u} \frac{1}{0.02} d u

Take the constant out:

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

= \frac{1}{0.02} \cdot \int e^{u} d u

Simplify

= 50 \cdot \int e^{u} d u

Use the common integral:

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

= 50 e^{u}

Substitute back

u = 0.02 x

= 50 e^{0.02 x}

Add a constant to the solution

= 50 e^{0.02 x} + C

\frac{d}{d x} (4 x ln (x))

\frac{d}{d x} (- cos (x) ln (sec (x) + tan (x)))

x^{3} - 6 x^{2} + 9 x + 1, x = 0.5, x = 4

f (x) = x^{2} + x, at x = - 4

\int_{0}^{1} e^{x} d x