What is the integral of 200e^{0.02x} ?

The integral of 200e^{0.02x} is 10000e^{0.02x}+C

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integral of 200e^{0.02x}

\int 200 e^{0.02 x} d x

10000 e^{0.02 x} + C

Solution steps

\int 200 e^{0.02 x} d x

Take the constant out:

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

= 200 \cdot \int e^{0.02 x} d x

Apply u-substitution

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

= 200 \cdot \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

= 200 \cdot \frac{1}{0.02} \cdot \int e^{u} d u

Simplify

= 200 \cdot 50 \cdot \int e^{u} d u

Use the common integral:

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

= 200 \cdot 50 e^{u}

Substitute back

u = 0.02 x

= 200 \cdot 50 e^{0.02 x}

Simplify

= 10000 e^{0.02 x}

Add a constant to the solution

= 10000 e^{0.02 x} + C

\int \frac{x ^{2}}{16 - x ^{6}} d x

\frac{d}{d x} (x sec^{2} (x))

y^{^{'}} = \frac{x ^{3} + y ^{3}}{x y ^{2}}

\frac{d y}{d x} y = 11 x

f (x) = - x^{4} + 4 x^{3} + 8 x^{2} + 5, at x = 2

What is the integral of 200e^{0.02x} ?
The integral of 200e^{0.02x} is 10000e^{0.02x}+C