Calculus Examples

$\frac{d y}{d x} = 2 x^{2} - \frac{x}{2} + 1$ , $y (1) = 2$

Step 1

Rewrite the equation.

$d y = (2 x^{2} - \frac{x}{2} + 1) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 2 x^{2} - \frac{x}{2} + 1 d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 2 x^{2} - \frac{x}{2} + 1 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 2 x^{2} d x + \int - \frac{x}{2} d x + \int d x$

Step 2.3.2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$y + C_{1} = 2 \int x^{2} d x + \int - \frac{x}{2} d x + \int d x$

Step 2.3.3

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

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

Step 2.3.4

Combine $\frac{1}{3}$ and $x^{3}$ .

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

Step 2.3.5

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 2.3.6

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

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

Step 2.3.7

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

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

Step 2.3.8

Apply the constant rule.

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

Step 2.3.9

Simplify.

$y + C_{1} = \frac{2 x^{3}}{3} - \frac{x^{2}}{4} + x + C_{5}$

Step 2.3.10

Reorder terms.

$y + C_{1} = \frac{2}{3} x^{3} - \frac{1}{4} x^{2} + x + C_{5}$

Step 2.4

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

$y = \frac{2}{3} x^{3} - \frac{1}{4} x^{2} + x + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $x$ and $2$ for $y$ in $y = \frac{2}{3} x^{3} - \frac{1}{4} x^{2} + x + K$ .

$2 = \frac{2}{3} \cdot 1^{3} - \frac{1}{4} \cdot 1^{2} + 1 + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\frac{2}{3} \cdot 1^{3} - \frac{1}{4} \cdot 1^{2} + 1 + K = 2$ .

$\frac{2}{3} \cdot 1^{3} - \frac{1}{4} \cdot 1^{2} + 1 + K = 2$

Step 4.2

Simplify $\frac{2}{3} \cdot 1^{3} - \frac{1}{4} \cdot 1^{2} + 1 + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

One to any power is one.

$\frac{2}{3} \cdot 1 - \frac{1}{4} \cdot 1^{2} + 1 + K = 2$

Step 4.2.1.2

Multiply $\frac{2}{3}$ by $1$ .

$\frac{2}{3} - \frac{1}{4} \cdot 1^{2} + 1 + K = 2$

Step 4.2.1.3

One to any power is one.

$\frac{2}{3} - \frac{1}{4} \cdot 1 + 1 + K = 2$

Step 4.2.1.4

Multiply $- 1$ by $1$ .

$\frac{2}{3} - \frac{1}{4} + 1 + K = 2$

Step 4.2.2

To write $\frac{2}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{2}{3} \cdot \frac{4}{4} - \frac{1}{4} + 1 + K = 2$

Step 4.2.3

To write $- \frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2}{3} \cdot \frac{4}{4} - \frac{1}{4} \cdot \frac{3}{3} + 1 + K = 2$

Step 4.2.4

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.2.4.1

Multiply $\frac{2}{3}$ by $\frac{4}{4}$ .

$\frac{2 \cdot 4}{3 \cdot 4} - \frac{1}{4} \cdot \frac{3}{3} + 1 + K = 2$

Step 4.2.4.2

Multiply $3$ by $4$ .

$\frac{2 \cdot 4}{12} - \frac{1}{4} \cdot \frac{3}{3} + 1 + K = 2$

Step 4.2.4.3

Multiply $\frac{1}{4}$ by $\frac{3}{3}$ .

$\frac{2 \cdot 4}{12} - \frac{3}{4 \cdot 3} + 1 + K = 2$

Step 4.2.4.4

Multiply $4$ by $3$ .

$\frac{2 \cdot 4}{12} - \frac{3}{12} + 1 + K = 2$

Step 4.2.5

Combine the numerators over the common denominator.

$\frac{2 \cdot 4 - 3}{12} + 1 + K = 2$

Step 4.2.6

Simplify the numerator.

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Step 4.2.6.1

Multiply $2$ by $4$ .

$\frac{8 - 3}{12} + 1 + K = 2$

Step 4.2.6.2

Subtract $3$ from $8$ .

$\frac{5}{12} + 1 + K = 2$

Step 4.2.7

Write $1$ as a fraction with a common denominator.

$\frac{5}{12} + \frac{12}{12} + K = 2$

Step 4.2.8

Combine the numerators over the common denominator.

$\frac{5 + 12}{12} + K = 2$

Step 4.2.9

Add $5$ and $12$ .

$\frac{17}{12} + K = 2$

Step 4.3

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

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Step 4.3.1

Subtract $\frac{17}{12}$ from both sides of the equation.

$K = 2 - \frac{17}{12}$

Step 4.3.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$K = 2 \cdot \frac{12}{12} - \frac{17}{12}$

Step 4.3.3

Combine $2$ and $\frac{12}{12}$ .

$K = \frac{2 \cdot 12}{12} - \frac{17}{12}$

Step 4.3.4

Combine the numerators over the common denominator.

$K = \frac{2 \cdot 12 - 17}{12}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Multiply $2$ by $12$ .

$K = \frac{24 - 17}{12}$

Step 4.3.5.2

Subtract $17$ from $24$ .

$K = \frac{7}{12}$

Step 5

Substitute $\frac{7}{12}$ for $K$ in $y = \frac{2}{3} x^{3} - \frac{1}{4} x^{2} + x + K$ and simplify.

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Step 5.1

Substitute $\frac{7}{12}$ for $K$ .

$y = \frac{2}{3} x^{3} - \frac{1}{4} x^{2} + x + \frac{7}{12}$

Step 5.2

Simplify each term.

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Step 5.2.1

Combine $\frac{2}{3}$ and $x^{3}$ .

$y = \frac{2 x^{3}}{3} - \frac{1}{4} x^{2} + x + \frac{7}{12}$

Step 5.2.2

Combine $x^{2}$ and $\frac{1}{4}$ .

$y = \frac{2 x^{3}}{3} - \frac{x^{2}}{4} + x + \frac{7}{12}$