Calculus Examples

$\frac{d y}{d x} = 4 x^{3} - 9 x^{2} + x$

Step 1

Rewrite the equation.

$d y = (4 x^{3} - 9 x^{2} + x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 4 x^{3} - 9 x^{2} + x d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 4 x^{3} - 9 x^{2} + x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 4 x^{3} d x + \int - 9 x^{2} d x + \int x d x$

Step 2.3.2

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

$y + C_{1} = 4 \int x^{3} d x + \int - 9 x^{2} d x + \int x d x$

Step 2.3.3

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

$y + C_{1} = 4 (\frac{1}{4} x^{4} + C_{2}) + \int - 9 x^{2} d x + \int x d x$

Step 2.3.4

Since $- 9$ is constant with respect to $x$ , move $- 9$ out of the integral.

$y + C_{1} = 4 (\frac{1}{4} x^{4} + C_{2}) - 9 \int x^{2} d x + \int x d x$

Step 2.3.5

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

$y + C_{1} = 4 (\frac{1}{4} x^{4} + C_{2}) - 9 (\frac{1}{3} x^{3} + C_{3}) + \int x d x$

Step 2.3.6

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

$y + C_{1} = 4 (\frac{1}{4} x^{4} + C_{2}) - 9 (\frac{1}{3} x^{3} + C_{3}) + \frac{1}{2} x^{2} + C_{4}$

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 $x^{4}$ .

$y + C_{1} = 4 (\frac{x^{4}}{4} + C_{2}) - 9 (\frac{1}{3} x^{3} + C_{3}) + \frac{1}{2} x^{2} + C_{4}$

Step 2.3.7.1.2

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

$y + C_{1} = 4 (\frac{x^{4}}{4} + C_{2}) - 9 (\frac{x^{3}}{3} + C_{3}) + \frac{1}{2} x^{2} + C_{4}$

Step 2.3.7.2

Simplify.

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

Step 2.4

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

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