Calculus Examples

$\frac{d y}{d x} = 4 \cdot x^{3} - 3 \cdot x^{2} + x$ , $y (1) = 0$

Step 1

Rewrite the equation.

$d y = (4 \cdot x^{3} - 3 \cdot x^{2} + x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 4 \cdot x^{3} - 3 \cdot x^{2} + x d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 4 \cdot x^{3} - 3 \cdot x^{2} + x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 4 \cdot x^{3} d x + \int - 3 \cdot x^{2} d x + \int x d x$

Step 2.3.2

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

$y + C_{1} = 4 \int x^{3} d x + \int - 3 \cdot x^{2} d x + \int x d x$

Step 2.3.3

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

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

Step 2.3.4

Since $- 3$ is constant with respect to $x$ , move $- 3$ out of the integral.

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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Step 2.3.7.1

Simplify.

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Step 2.3.7.1.1

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

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

Step 2.3.7.1.2

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

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

Step 2.3.7.2

Simplify.

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

Step 2.4

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

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

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $x$ and $0$ for $y$ in $y = x^{4} - x^{3} + \frac{1}{2} x^{2} + K$ .

$0 = 1^{4} - 1^{3} + \frac{1}{2} \cdot 1^{2} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $1^{4} - 1^{3} + \frac{1}{2} \cdot 1^{2} + K = 0$ .

$1^{4} - 1^{3} + \frac{1}{2} \cdot 1^{2} + K = 0$

Step 4.2

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

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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.

$1 - 1^{3} + \frac{1}{2} \cdot 1^{2} + K = 0$

Step 4.2.1.2

One to any power is one.

$1 - 1 \cdot 1 + \frac{1}{2} \cdot 1^{2} + K = 0$

Step 4.2.1.3

Multiply $- 1$ by $1$ .

$1 - 1 + \frac{1}{2} \cdot 1^{2} + K = 0$

Step 4.2.1.4

One to any power is one.

$1 - 1 + \frac{1}{2} \cdot 1 + K = 0$

Step 4.2.1.5

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

$1 - 1 + \frac{1}{2} + K = 0$

Step 4.2.2

Simplify by subtracting numbers.

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Step 4.2.2.1

Subtract $1$ from $1$ .

$0 + \frac{1}{2} + K = 0$

Step 4.2.2.2

Add $0$ and $\frac{1}{2}$ .

$\frac{1}{2} + K = 0$

Step 4.3

Subtract $\frac{1}{2}$ from both sides of the equation.

$K = - \frac{1}{2}$

Step 5

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

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Step 5.1

Substitute $- \frac{1}{2}$ for $K$ .

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

Step 5.2

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

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