Calculus Examples

$\frac{d y}{d x} = 8 x + 8$ , $y (- 1) = 0$

Step 1

Rewrite the equation.

$d y = (8 x + 8) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 8 x + 8 d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 8 x + 8 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 8 x d x + \int 8 d x$

Step 2.3.2

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$y + C_{1} = 8 \int x d x + \int 8 d x$

Step 2.3.3

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

$y + C_{1} = 8 (\frac{1}{2} x^{2} + C_{2}) + \int 8 d x$

Step 2.3.4

Apply the constant rule.

$y + C_{1} = 8 (\frac{1}{2} x^{2} + C_{2}) + 8 x + C_{3}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{2}$ and $x^{2}$ .

$y + C_{1} = 8 (\frac{x^{2}}{2} + C_{2}) + 8 x + C_{3}$

Step 2.3.5.2

Simplify.

$y + C_{1} = 4 x^{2} + 8 x + C_{4}$

Step 2.4

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

$y = 4 x^{2} + 8 x + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $- 1$ for $x$ and $0$ for $y$ in $y = 4 x^{2} + 8 x + K$ .

$0 = 4 {(- 1)}^{2} + 8 \cdot - 1 + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $4 {(- 1)}^{2} + 8 \cdot - 1 + K = 0$ .

$4 {(- 1)}^{2} + 8 \cdot - 1 + K = 0$

Step 4.2

Simplify $4 {(- 1)}^{2} + 8 \cdot - 1 + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $- 1$ to the power of $2$ .

$4 \cdot 1 + 8 \cdot - 1 + K = 0$

Step 4.2.1.2

Multiply $4$ by $1$ .

$4 + 8 \cdot - 1 + K = 0$

Step 4.2.1.3

Multiply $8$ by $- 1$ .

$4 - 8 + K = 0$

Step 4.2.2

Subtract $8$ from $4$ .

$- 4 + K = 0$

Step 4.3

Add $4$ to both sides of the equation.

$K = 4$

Step 5

Substitute $4$ for $K$ in $y = 4 x^{2} + 8 x + K$ and simplify.

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Step 5.1

Substitute $4$ for $K$ .

$y = 4 x^{2} + 8 x + 4$