Calculus Examples

$\frac{d y}{d x} = x^{2}$ , $y (0) = 4$

Step 1

Rewrite the equation.

$d y = x^{2} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int x^{2} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

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

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

Step 4

Solve for $K$ .

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Step 4.1

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

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

Step 4.2

Simplify $\frac{1}{3} \cdot 0^{3} + K$ .

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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 $0$ .

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

Step 4.2.1.2

Multiply $\frac{1}{3}$ by $0$ .

$0 + K = 4$

Step 4.2.2

Add $0$ and $K$ .

$K = 4$

Step 5

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

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Step 5.1

Substitute $4$ for $K$ .

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

Step 5.2

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

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