Calculus Examples

$\frac{d f}{d x} = 4 x^{2} - 6 e^{x}$ , $f (0) = - 5$

Step 1

Rewrite the equation.

$d f = (4 x^{2} - 6 e^{x}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d f = \int 4 x^{2} - 6 e^{x} d x$

Step 2.2

Apply the constant rule.

$f + C_{1} = \int 4 x^{2} - 6 e^{x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$f + C_{1} = \int 4 x^{2} d x + \int - 6 e^{x} d x$

Step 2.3.2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$f + C_{1} = 4 \int x^{2} d x + \int - 6 e^{x} d x$

Step 2.3.3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$f + C_{1} = 4 (\frac{1}{3} x^{3} + C_{2}) + \int - 6 e^{x} d x$

Step 2.3.4

Since $- 6$ is constant with respect to $x$ , move $- 6$ out of the integral.

$f + C_{1} = 4 (\frac{1}{3} x^{3} + C_{2}) - 6 \int e^{x} d x$

Step 2.3.5

The integral of $e^{x}$ with respect to $x$ is $e^{x}$ .

$f + C_{1} = 4 (\frac{1}{3} x^{3} + C_{2}) - 6 (e^{x} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$f + C_{1} = 4 (\frac{1}{3}) x^{3} - 6 e^{x} + C_{4}$

Step 2.3.6.2

Combine $4$ and $\frac{1}{3}$ .

$f + C_{1} = \frac{4}{3} x^{3} - 6 e^{x} + C_{4}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$f = \frac{4}{3} x^{3} - 6 e^{x} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $- 5$ for $f$ in $f = \frac{4}{3} x^{3} - 6 e^{x} + K$ .

$- 5 = \frac{4}{3} \cdot 0^{3} - 6 e^{0} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\frac{4}{3} \cdot 0^{3} - 6 e^{0} + K = - 5$ .

$\frac{4}{3} \cdot 0^{3} - 6 e^{0} + K = - 5$

Step 4.2

Simplify $\frac{4}{3} \cdot 0^{3} - 6 e^{0} + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raising $0$ to any positive power yields $0$ .

$\frac{4}{3} \cdot 0 - 6 e^{0} + K = - 5$

Step 4.2.1.2

Multiply $\frac{4}{3}$ by $0$ .

$0 - 6 e^{0} + K = - 5$

Step 4.2.1.3

Anything raised to $0$ is $1$ .

$0 - 6 \cdot 1 + K = - 5$

Step 4.2.1.4

Multiply $- 6$ by $1$ .

$0 - 6 + K = - 5$

Step 4.2.2

Subtract $6$ from $0$ .

$- 6 + K = - 5$

Step 4.3

Move all terms not containing $K$ to the right side of the equation.

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Step 4.3.1

Add $6$ to both sides of the equation.

$K = - 5 + 6$

Step 4.3.2

Add $- 5$ and $6$ .

$K = 1$

Step 5

Substitute $1$ for $K$ in $f = \frac{4}{3} x^{3} - 6 e^{x} + K$ and simplify.

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Step 5.1

Substitute $1$ for $K$ .

$f = \frac{4}{3} x^{3} - 6 e^{x} + 1$

Step 5.2

Combine $\frac{4}{3}$ and $x^{3}$ .

$f = \frac{4 x^{3}}{3} - 6 e^{x} + 1$