Calculus Examples

$\int_{0}^{\infty} x e^{- 2 x} d x$

Step 1

Write the integral as a limit as $t$ approaches $\infty$ .

$lim_{t \to \infty} \int_{0}^{t} x e^{- 2 x} d x$

Step 2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = e^{- 2 x}$ .

$lim_{t \to \infty} x (- \frac{1}{2} e^{- 2 x})]_{0}^{t} - \int_{0}^{t} - \frac{1}{2} e^{- 2 x} d x$

Step 3

Simplify.

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Step 3.1

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

$lim_{t \to \infty} x (- \frac{e^{- 2 x}}{2})]_{0}^{t} - \int_{0}^{t} - \frac{1}{2} e^{- 2 x} d x$

Step 3.2

Combine $x$ and $\frac{e^{- 2 x}}{2}$ .

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} - \int_{0}^{t} - \frac{1}{2} e^{- 2 x} d x$

Step 4

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} - (- \frac{1}{2} \int_{0}^{t} e^{- 2 x} d x)$

Step 5

Simplify.

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Step 5.1

Multiply $- 1$ by $- 1$ .

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + 1 (\frac{1}{2} \int_{0}^{t} e^{- 2 x} d x)$

Step 5.2

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + \frac{1}{2} \int_{0}^{t} e^{- 2 x} d x$

Step 6

Let $u = - 2 x$ . 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 6.1

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

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Step 6.1.1

Differentiate $- 2 x$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

$- 2 \cdot 1$

Step 6.1.4

Multiply $- 2$ by $1$ .

$- 2$

Step 6.2

Substitute the lower limit in for $x$ in $u = - 2 x$ .

$u_{lower} = - 2 \cdot 0$

Step 6.3

Multiply $- 2$ by $0$ .

$u_{lower} = 0$

Step 6.4

Substitute the upper limit in for $x$ in $u = - 2 x$ .

$u_{upper} = - 2 t$

Step 6.5

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = - 2 t$

Step 6.6

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + \frac{1}{2} \int_{0}^{- 2 t} e^{u} \frac{1}{- 2} d u$

Step 7

Simplify.

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Step 7.1

Move the negative in front of the fraction.

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + \frac{1}{2} \int_{0}^{- 2 t} e^{u} (- \frac{1}{2}) d u$

Step 7.2

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + \frac{1}{2} \int_{0}^{- 2 t} - \frac{e^{u}}{2} d u$

Step 8

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + \frac{1}{2} (- \int_{0}^{- 2 t} \frac{e^{u}}{2} d u)$

Step 9

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} + \frac{1}{2} (- (\frac{1}{2} \int_{0}^{- 2 t} e^{u} d u))$

Step 10

Simplify.

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Step 10.1

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} - \frac{1}{2 \cdot 2} \int_{0}^{- 2 t} e^{u} d u$

Step 10.2

Multiply $2$ by $2$ .

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} - \frac{1}{4} \int_{0}^{- 2 t} e^{u} d u$

Step 11

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

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} - \frac{1}{4} e^{u}]_{0}^{- 2 t}$

Step 12

Combine $e^{u}]_{0}^{- 2 t}$ and $\frac{1}{4}$ .

$lim_{t \to \infty} - \frac{x e^{- 2 x}}{2}]_{0}^{t} - \frac{e^{u}]_{0}^{- 2 t}}{4}$

Step 13

Substitute and simplify.

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Step 13.1

Evaluate $- \frac{x e^{- 2 x}}{2}$ at $t$ and at $0$ .

$lim_{t \to \infty} (- \frac{t e^{- 2 t}}{2}) + \frac{0 e^{- 2 \cdot 0}}{2} - \frac{e^{u}]_{0}^{- 2 t}}{4}$

Step 13.2

Evaluate $e^{u}$ at $- 2 t$ and at $0$ .

$lim_{t \to \infty} (- \frac{t e^{- 2 t}}{2}) + \frac{0 e^{- 2 \cdot 0}}{2} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3

Simplify.

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Step 13.3.1

Multiply $- 2$ by $0$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{0 e^{0}}{2} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.2

Anything raised to $0$ is $1$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{0 \cdot 1}{2} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.3

Multiply $0$ by $1$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{0}{2} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.4

Cancel the common factor of $0$ and $2$ .

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Step 13.3.4.1

Factor $2$ out of $0$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{2 (0)}{2} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.4.2

Cancel the common factors.

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Step 13.3.4.2.1

Factor $2$ out of $2$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{2 \cdot 0}{2 \cdot 1} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.4.2.2

Cancel the common factor.

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{2 \cdot 0}{2 \cdot 1} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.4.2.3

Rewrite the expression.

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + \frac{0}{1} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.4.2.4

Divide $0$ by $1$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} + 0 - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.5

Add $- \frac{t e^{- 2 t}}{2}$ and $0$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} - \frac{(e^{- 2 t}) - e^{0}}{4}$

Step 13.3.6

Anything raised to $0$ is $1$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} - \frac{e^{- 2 t} - 1 \cdot 1}{4}$

Step 13.3.7

Multiply $- 1$ by $1$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} - \frac{e^{- 2 t} - 1}{4}$

Step 13.3.8

To write $- \frac{t e^{- 2 t}}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t}}{2} \cdot \frac{2}{2} - \frac{e^{- 2 t} - 1}{4}$

Step 13.3.9

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 13.3.9.1

Multiply $\frac{t e^{- 2 t}}{2}$ by $\frac{2}{2}$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t} \cdot 2}{2 \cdot 2} - \frac{e^{- 2 t} - 1}{4}$

Step 13.3.9.2

Multiply $2$ by $2$ .

$lim_{t \to \infty} - \frac{t e^{- 2 t} \cdot 2}{4} - \frac{e^{- 2 t} - 1}{4}$

Step 13.3.10

Combine the numerators over the common denominator.

$lim_{t \to \infty} \frac{- t e^{- 2 t} \cdot 2 - (e^{- 2 t} - 1)}{4}$

Step 13.3.11

Multiply $2$ by $- 1$ .

$lim_{t \to \infty} \frac{- 2 t e^{- 2 t} - (e^{- 2 t} - 1)}{4}$

Step 14

Simplify.

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Step 14.1

Factor $- 1$ out of $- 2 t e^{- 2 t}$ .

$lim_{t \to \infty} \frac{- (2 t e^{- 2 t}) - (e^{- 2 t} - 1)}{4}$

Step 14.2

Factor $- 1$ out of $- (2 t e^{- 2 t}) - (e^{- 2 t} - 1)$ .

$lim_{t \to \infty} \frac{- (2 t e^{- 2 t} + e^{- 2 t} - 1)}{4}$

Step 14.3

Rewrite $- (2 t e^{- 2 t} + e^{- 2 t} - 1)$ as $- 1 (2 t e^{- 2 t} + e^{- 2 t} - 1)$ .

$lim_{t \to \infty} \frac{- 1 (2 t e^{- 2 t} + e^{- 2 t} - 1)}{4}$

Step 14.4

Move the negative in front of the fraction.

$lim_{t \to \infty} - \frac{2 t e^{- 2 t} + e^{- 2 t} - 1}{4}$

Step 15

Evaluate the limit.

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Step 15.1

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

$- lim_{t \to \infty} \frac{2 t e^{- 2 t} + e^{- 2 t} - 1}{4}$

Step 15.2

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{4} lim_{t \to \infty} 2 t e^{- 2 t} + e^{- 2 t} - 1$

Step 15.3

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $\infty$ .

$- \frac{1}{4} (lim_{t \to \infty} 2 t e^{- 2 t} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.4

Move the term $2$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{4} (2 lim_{t \to \infty} t e^{- 2 t} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.5

Rewrite $t e^{- 2 t}$ as $\frac{t}{e^{2 t}}$ .

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{t}{e^{2 t}} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6

Apply L'Hospital's rule.

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Step 15.6.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 15.6.1.1

Take the limit of the numerator and the limit of the denominator.

$- \frac{1}{4} (2 \frac{lim_{t \to \infty} t}{lim_{t \to \infty} e^{2 t}} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$- \frac{1}{4} (2 \frac{\infty}{lim_{t \to \infty} e^{2 t}} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.1.3

Since the exponent $2 t$ approaches $\infty$ , the quantity $e^{2 t}$ approaches $\infty$ .

$- \frac{1}{4} (2 \frac{\infty}{\infty} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.1.4

Infinity divided by infinity is undefined.

Undefined

$- \frac{1}{4} (2 \frac{\infty}{\infty} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.2

Since $\frac{\infty}{\infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{t \to \infty} \frac{t}{e^{2 t}} = lim_{t \to \infty} \frac{\frac{d}{d t} [t]}{\frac{d}{d t} [e^{2 t}]}$

Step 15.6.3

Find the derivative of the numerator and denominator.

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Step 15.6.3.1

Differentiate the numerator and denominator.

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{\frac{d}{d t} [t]}{\frac{d}{d t} [e^{2 t}]} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.2

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

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{\frac{d}{d t} [e^{2 t}]} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 2 t$ .

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Step 15.6.3.3.1

To apply the Chain Rule, set $u$ as $2 t$ .

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d t} [2 t]} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{e^{u} \frac{d}{d t} [2 t]} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.3.3

Replace all occurrences of $u$ with $2 t$ .

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{e^{2 t} \frac{d}{d t} [2 t]} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.4

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

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{e^{2 t} \cdot 2 \frac{d}{d t} [t]} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.5

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

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{e^{2 t} \cdot 2 \cdot 1} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.6

Multiply $2$ by $1$ .

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{e^{2 t} \cdot 2} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.6.3.7

Move $2$ to the left of $e^{2 t}$ .

$- \frac{1}{4} (2 lim_{t \to \infty} \frac{1}{2 e^{2 t}} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.7

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{4} (2 (\frac{1}{2}) lim_{t \to \infty} \frac{1}{e^{2 t}} + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.8

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{e^{2 t}}$ approaches $0$ .

$- \frac{1}{4} (2 (\frac{1}{2}) \cdot 0 + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.9

Multiply $\frac{1}{2}$ by $0$ .

$- \frac{1}{4} (2 \cdot 0 + lim_{t \to \infty} e^{- 2 t} - lim_{t \to \infty} 1)$

Step 15.10

Since the exponent $- 2 t$ approaches $- \infty$ , the quantity $e^{- 2 t}$ approaches $0$ .

$- \frac{1}{4} (2 \cdot 0 + 0 - lim_{t \to \infty} 1)$

Step 15.11

Evaluate the limit.

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Step 15.11.1

Evaluate the limit of $1$ which is constant as $t$ approaches $\infty$ .

$- \frac{1}{4} (2 \cdot 0 + 0 - 1 \cdot 1)$

Step 15.11.2

Simplify the answer.

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Step 15.11.2.1

Simplify each term.

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Step 15.11.2.1.1

Multiply $2$ by $0$ .

$- \frac{1}{4} (0 + 0 - 1 \cdot 1)$

Step 15.11.2.1.2

Multiply $- 1$ by $1$ .

$- \frac{1}{4} (0 + 0 - 1)$

Step 15.11.2.2

Add $0$ and $0$ .

$- \frac{1}{4} (0 - 1)$

Step 15.11.2.3

Subtract $1$ from $0$ .

$- \frac{1}{4} \cdot - 1$

Step 15.11.2.4

Multiply $- \frac{1}{4} \cdot - 1$ .

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Step 15.11.2.4.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{4})$

Step 15.11.2.4.2

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

$\frac{1}{4}$

Step 16

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{4}$

Decimal Form:

$0.25$