Calculus Examples

$\frac{d F}{d x} = - 5 x^{2} - 5 e^{x}$ , $F (0) = - 9$

Step 1

Rewrite the equation.

$d F = (- 5 x^{2} - 5 e^{x}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d F = \int - 5 x^{2} - 5 e^{x} d x$

Step 2.2

Apply the constant rule.

$F + C_{1} = \int - 5 x^{2} - 5 e^{x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$F + C_{1} = \int - 5 x^{2} d x + \int - 5 e^{x} d x$

Step 2.3.2

Since $- 5$ is constant with respect to $x$ , move $- 5$ out of the integral.

$F + C_{1} = - 5 \int x^{2} d x + \int - 5 e^{x} d x$

Step 2.3.3

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

$F + C_{1} = - 5 (\frac{1}{3} x^{3} + C_{2}) + \int - 5 e^{x} d x$

Step 2.3.4

Since $- 5$ is constant with respect to $x$ , move $- 5$ out of the integral.

$F + C_{1} = - 5 (\frac{1}{3} x^{3} + C_{2}) - 5 \int e^{x} d x$

Step 2.3.5

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

$F + C_{1} = - 5 (\frac{1}{3} x^{3} + C_{2}) - 5 (e^{x} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$F + C_{1} = - 5 (\frac{1}{3}) x^{3} - 5 e^{x} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $- 5$ and $\frac{1}{3}$ .

$F + C_{1} = \frac{- 5}{3} x^{3} - 5 e^{x} + C_{4}$

Step 2.3.6.2.2

Move the negative in front of the fraction.

$F + C_{1} = - \frac{5}{3} x^{3} - 5 e^{x} + C_{4}$

Step 2.4

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

$F = - \frac{5}{3} x^{3} - 5 e^{x} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $- 9$ for $F$ in $F = - \frac{5}{3} x^{3} - 5 e^{x} + K$ .

$- 9 = - \frac{5}{3} \cdot 0^{3} - 5 e^{0} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $- \frac{5}{3} \cdot 0^{3} - 5 e^{0} + K = - 9$ .

$- \frac{5}{3} \cdot 0^{3} - 5 e^{0} + K = - 9$

Step 4.2

Simplify $- \frac{5}{3} \cdot 0^{3} - 5 e^{0} + K$ .

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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 $0$ .

$- \frac{5}{3} \cdot 0 - 5 e^{0} + K = - 9$

Step 4.2.1.2

Multiply $- \frac{5}{3} \cdot 0$ .

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Step 4.2.1.2.1

Multiply $0$ by $- 1$ .

$0 (\frac{5}{3}) - 5 e^{0} + K = - 9$

Step 4.2.1.2.2

Multiply $0$ by $\frac{5}{3}$ .

$0 - 5 e^{0} + K = - 9$

Step 4.2.1.3

Anything raised to $0$ is $1$ .

$0 - 5 \cdot 1 + K = - 9$

Step 4.2.1.4

Multiply $- 5$ by $1$ .

$0 - 5 + K = - 9$

Step 4.2.2

Subtract $5$ from $0$ .

$- 5 + K = - 9$

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 $5$ to both sides of the equation.

$K = - 9 + 5$

Step 4.3.2

Add $- 9$ and $5$ .

$K = - 4$

Step 5

Substitute $- 4$ for $K$ in $F = - \frac{5}{3} x^{3} - 5 e^{x} + K$ and simplify.

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Step 5.1

Substitute $- 4$ for $K$ .

$F = - \frac{5}{3} x^{3} - 5 e^{x} - 4$

Step 5.2

Simplify each term.

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Step 5.2.1

Combine $x^{3}$ and $\frac{5}{3}$ .

$F = - \frac{x^{3} \cdot 5}{3} - 5 e^{x} - 4$

Step 5.2.2

Move $5$ to the left of $x^{3}$ .

$F = - \frac{5 x^{3}}{3} - 5 e^{x} - 4$