Calculus Examples

$\int_{4}^{\infty} \frac{1}{{(x - 3)}^{\frac{3}{2}}} d x$

Step 1

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

$lim_{t \to \infty} \int_{4}^{t} \frac{1}{{(x - 3)}^{\frac{3}{2}}} d x$

Step 2

Let $u = x - 3$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = x - 3$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $x - 3$ .

$\frac{d}{d x} [x - 3]$

Step 2.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 2.1.3

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

$1 + \frac{d}{d x} [- 3]$

Step 2.1.4

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.1.5

Add $1$ and $0$ .

$1$

Step 2.2

Substitute the lower limit in for $x$ in $u = x - 3$ .

$u_{lower} = 4 - 3$

Step 2.3

Subtract $3$ from $4$ .

$u_{lower} = 1$

Step 2.4

Substitute the upper limit in for $x$ in $u = x - 3$ .

$u_{upper} = t - 3$

Step 2.5

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

$u_{lower} = 1$

$u_{upper} = t - 3$

Step 2.6

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

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

Step 3

Apply basic rules of exponents.

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Step 3.1

Move $u^{\frac{3}{2}}$ out of the denominator by raising it to the $- 1$ power.

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

Step 3.2

Multiply the exponents in ${(u^{\frac{3}{2}})}^{- 1}$ .

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Step 3.2.1

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

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

Step 3.2.2

Multiply $\frac{3}{2} \cdot - 1$ .

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Step 3.2.2.1

Combine $\frac{3}{2}$ and $- 1$ .

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

Step 3.2.2.2

Multiply $3$ by $- 1$ .

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

Step 3.2.3

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

Substitute and simplify.

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Step 5.1

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

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

Step 5.2

Simplify.

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Step 5.2.1

One to any power is one.

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

Step 5.2.2

Multiply $2$ by $1$ .

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

Step 6

Evaluate the limit.

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Step 6.1

Evaluate the limit.

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Step 6.1.1

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

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

Step 6.1.2

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

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

Step 6.1.3

Simplify the limit argument.

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Step 6.1.3.1

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

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

Step 6.1.3.2

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

$- 2 lim_{t \to \infty} \frac{1}{\sqrt{t - 3}} + lim_{t \to \infty} 2$

Step 6.2

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

$- 2 \cdot 0 + lim_{t \to \infty} 2$

Step 6.3

Evaluate the limit.

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Step 6.3.1

Evaluate the limit of $2$ which is constant as $t$ approaches $\infty$ .

$- 2 \cdot 0 + 2$

Step 6.3.2

Simplify the answer.

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Step 6.3.2.1

Multiply $- 2$ by $0$ .

$0 + 2$

Step 6.3.2.2

Add $0$ and $2$ .

$2$