Calculus Examples

$\int_{1}^{\infty} \frac{x^{2}}{{(x^{3} + 2)}^{2}} d x$

Step 1

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

$lim_{t \to \infty} \int_{1}^{t} \frac{x^{2}}{{(x^{3} + 2)}^{2}} d x$

Step 2

Let $u = x^{3} + 2$ . Then $d u = 3 x^{2} d x$ , so $\frac{1}{3} d u = x^{2} d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = x^{3} + 2$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $x^{3} + 2$ .

$\frac{d}{d x} [x^{3} + 2]$

Step 2.1.2

By the Sum Rule, the derivative of $x^{3} + 2$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [2]$

Step 2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2} + \frac{d}{d x} [2]$

Step 2.1.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$3 x^{2} + 0$

Step 2.1.5

Add $3 x^{2}$ and $0$ .

$3 x^{2}$

Step 2.2

Substitute the lower limit in for $x$ in $u = x^{3} + 2$ .

$u_{lower} = 1^{3} + 2$

Step 2.3

Simplify.

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Step 2.3.1

One to any power is one.

$u_{lower} = 1 + 2$

Step 2.3.2

Add $1$ and $2$ .

$u_{lower} = 3$

Step 2.4

Substitute the upper limit in for $x$ in $u = x^{3} + 2$ .

$u_{upper} = t^{3} + 2$

Step 2.5

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

$u_{lower} = 3$

$u_{upper} = t^{3} + 2$

Step 2.6

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

$lim_{t \to \infty} \int_{3}^{t^{3} + 2} \frac{1}{u^{2}} \cdot \frac{1}{3} d u$

Step 3

Simplify.

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Step 3.1

Multiply $\frac{1}{u^{2}}$ by $\frac{1}{3}$ .

$lim_{t \to \infty} \int_{3}^{t^{3} + 2} \frac{1}{u^{2} \cdot 3} d u$

Step 3.2

Move $3$ to the left of $u^{2}$ .

$lim_{t \to \infty} \int_{3}^{t^{3} + 2} \frac{1}{3 u^{2}} d u$

Step 4

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$lim_{t \to \infty} \frac{1}{3} \int_{3}^{t^{3} + 2} \frac{1}{u^{2}} d u$

Step 5

Apply basic rules of exponents.

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Step 5.1

Move $u^{2}$ out of the denominator by raising it to the $- 1$ power.

$lim_{t \to \infty} \frac{1}{3} \int_{3}^{t^{3} + 2} {(u^{2})}^{- 1} d u$

Step 5.2

Multiply the exponents in ${(u^{2})}^{- 1}$ .

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Step 5.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$lim_{t \to \infty} \frac{1}{3} \int_{3}^{t^{3} + 2} u^{2 \cdot - 1} d u$

Step 5.2.2

Multiply $2$ by $- 1$ .

$lim_{t \to \infty} \frac{1}{3} \int_{3}^{t^{3} + 2} u^{- 2} d u$

Step 6

By the Power Rule, the integral of $u^{- 2}$ with respect to $u$ is $- u^{- 1}$ .

$lim_{t \to \infty} \frac{1}{3} - u^{- 1}]_{3}^{t^{3} + 2}$

Step 7

Combine $\frac{1}{3}$ and $- u^{- 1}]_{3}^{t^{3} + 2}$ .

$lim_{t \to \infty} \frac{- u^{- 1}]_{3}^{t^{3} + 2}}{3}$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $- u^{- 1}$ at $t^{3} + 2$ and at $3$ .

$lim_{t \to \infty} \frac{(- {(t^{3} + 2)}^{- 1}) + 3^{- 1}}{3}$

Step 8.2

Simplify.

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Step 8.2.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{t \to \infty} \frac{- {(t^{3} + 2)}^{- 1} + \frac{1}{3}}{3}$

Step 8.2.2

To write $- {(t^{3} + 2)}^{- 1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$lim_{t \to \infty} \frac{- {(t^{3} + 2)}^{- 1} \cdot \frac{3}{3} + \frac{1}{3}}{3}$

Step 8.2.3

Combine $- {(t^{3} + 2)}^{- 1}$ and $\frac{3}{3}$ .

$lim_{t \to \infty} \frac{\frac{- {(t^{3} + 2)}^{- 1} \cdot 3}{3} + \frac{1}{3}}{3}$

Step 8.2.4

Combine the numerators over the common denominator.

$lim_{t \to \infty} \frac{\frac{- {(t^{3} + 2)}^{- 1} \cdot 3 + 1}{3}}{3}$

Step 8.2.5

Multiply $3$ by $- 1$ .

$lim_{t \to \infty} \frac{\frac{- 3 {(t^{3} + 2)}^{- 1} + 1}{3}}{3}$

Step 8.2.6

Rewrite $\frac{\frac{- 3 {(t^{3} + 2)}^{- 1} + 1}{3}}{3}$ as a product.

$lim_{t \to \infty} \frac{- 3 {(t^{3} + 2)}^{- 1} + 1}{3} \cdot \frac{1}{3}$

Step 8.2.7

Multiply $\frac{- 3 {(t^{3} + 2)}^{- 1} + 1}{3}$ by $\frac{1}{3}$ .

$lim_{t \to \infty} \frac{- 3 {(t^{3} + 2)}^{- 1} + 1}{3 \cdot 3}$

Step 8.2.8

Multiply $3$ by $3$ .

$lim_{t \to \infty} \frac{- 3 {(t^{3} + 2)}^{- 1} + 1}{9}$

Step 9

Simplify.

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Step 9.1

Factor $- 1$ out of $- 3 {(t^{3} + 2)}^{- 1}$ .

$lim_{t \to \infty} \frac{- (3 {(t^{3} + 2)}^{- 1}) + 1}{9}$

Step 9.2

Rewrite $1$ as $- 1 (- 1)$ .

$lim_{t \to \infty} \frac{- (3 {(t^{3} + 2)}^{- 1}) - 1 (- 1)}{9}$

Step 9.3

Factor $- 1$ out of $- (3 {(t^{3} + 2)}^{- 1}) - 1 (- 1)$ .

$lim_{t \to \infty} \frac{- (3 {(t^{3} + 2)}^{- 1} - 1)}{9}$

Step 9.4

Rewrite $- (3 {(t^{3} + 2)}^{- 1} - 1)$ as $- 1 (3 {(t^{3} + 2)}^{- 1} - 1)$ .

$lim_{t \to \infty} \frac{- 1 (3 {(t^{3} + 2)}^{- 1} - 1)}{9}$

Step 9.5

Move the negative in front of the fraction.

$lim_{t \to \infty} - \frac{3 {(t^{3} + 2)}^{- 1} - 1}{9}$

Step 10

Evaluate the limit.

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Step 10.1

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

$- lim_{t \to \infty} \frac{3 {(t^{3} + 2)}^{- 1} - 1}{9}$

Step 10.2

Move the term $\frac{1}{9}$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{9} lim_{t \to \infty} 3 {(t^{3} + 2)}^{- 1} - 1$

Step 10.3

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

$- \frac{1}{9} (lim_{t \to \infty} 3 {(t^{3} + 2)}^{- 1} - lim_{t \to \infty} 1)$

Step 10.4

Move the term $3$ outside of the limit because it is constant with respect to $t$ .

$- \frac{1}{9} (3 lim_{t \to \infty} {(t^{3} + 2)}^{- 1} - lim_{t \to \infty} 1)$

Step 10.5

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{9} (3 lim_{t \to \infty} \frac{1}{t^{3} + 2} - lim_{t \to \infty} 1)$

Step 10.6

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{t^{3} + 2}$ approaches $0$ .

$- \frac{1}{9} (3 \cdot 0 - lim_{t \to \infty} 1)$

Step 10.7

Evaluate the limit.

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Step 10.7.1

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

$- \frac{1}{9} (3 \cdot 0 - 1 \cdot 1)$

Step 10.7.2

Simplify the answer.

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Step 10.7.2.1

Simplify each term.

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Step 10.7.2.1.1

Multiply $3$ by $0$ .

$- \frac{1}{9} (0 - 1 \cdot 1)$

Step 10.7.2.1.2

Multiply $- 1$ by $1$ .

$- \frac{1}{9} (0 - 1)$

Step 10.7.2.2

Subtract $1$ from $0$ .

$- \frac{1}{9} \cdot - 1$

Step 10.7.2.3

Multiply $- \frac{1}{9} \cdot - 1$ .

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Step 10.7.2.3.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{9})$

Step 10.7.2.3.2

Multiply $\frac{1}{9}$ by $1$ .

$\frac{1}{9}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{9}$

Decimal Form:

$0.\overline{1}$