Calculus Examples

$e^{2 x - 1} y^{- 2}$

Step 1

Write $e^{2 x - 1} y^{- 2}$ as a function.

$f (x) = e^{2 x - 1} y^{- 2}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int e^{2 x - 1} y^{- 2} d x$

Step 4

Since $y^{- 2}$ is constant with respect to $x$ , move $y^{- 2}$ out of the integral.

$y^{- 2} \int e^{2 x - 1} d x$

Step 5

Let $u = 2 x - 1$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = 2 x - 1$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $2 x - 1$ .

$\frac{d}{d x} [2 x - 1]$

Step 5.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 5.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 5.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 5.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [- 1]$

Step 5.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 1]$

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 + 0$

Step 5.1.4.2

Add $2$ and $0$ .

$2$

Step 5.2

Rewrite the problem using $u$ and $d u$ .

$y^{- 2} \int e^{u} \frac{1}{2} d u$

Step 6

Combine $e^{u}$ and $\frac{1}{2}$ .

$y^{- 2} \int \frac{e^{u}}{2} d u$

Step 7

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

$y^{- 2} (\frac{1}{2} \int e^{u} d u)$

Step 8

Simplify.

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Step 8.1

Combine $\frac{1}{2}$ and $y^{- 2}$ .

$\frac{y^{- 2}}{2} \int e^{u} d u$

Step 8.2

Move $y^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 y^{2}} \int e^{u} d u$

Step 9

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$\frac{1}{2 y^{2}} (e^{u} + C)$

Step 10

Simplify.

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Step 10.1

Simplify.

$\frac{1}{2 y^{2}} e^{u} + C$

Step 10.2

Combine $\frac{1}{2 y^{2}}$ and $e^{u}$ .

$\frac{e^{u}}{2 y^{2}} + C$

Step 11

Replace all occurrences of $u$ with $2 x - 1$ .

$\frac{e^{2 x - 1}}{2 y^{2}} + C$

Step 12

The answer is the antiderivative of the function $f (x) = e^{2 x - 1} y^{- 2}$ .

$F (x) =$ $\frac{e^{2 x - 1}}{2 y^{2}} + C$