Calculus Examples

$\int_{- \infty}^{0} \frac{1}{16 + x^{2}} d x$

Step 1

This integral could not be completed using u-substitution. Mathway will use another method.

Step 2

Write the integral as a limit as

$t$ approaches

$- \infty$ .

$lim_{t \to - \infty} \int_{t}^{0} \frac{1}{16 + x^{2}} d x$

Step 3

Rewrite

$16$ as

$4^{2}$ .

$lim_{t \to - \infty} \int_{t}^{0} \frac{1}{4^{2} + x^{2}} d x$

Step 4

The integral of

$\frac{1}{4^{2} + x^{2}}$ with respect to

$x$ is

$\frac{1}{4} \arctan (\frac{x}{4})]_{t}^{0}$ .

$lim_{t \to - \infty} \frac{1}{4} \arctan (\frac{x}{4})]_{t}^{0}$

Step 5

Simplify the answer.

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Step 5.1

Combine

$\frac{1}{4}$ and

$\arctan (\frac{x}{4})$ .

$lim_{t \to - \infty} \frac{\arctan (\frac{x}{4})}{4}]_{t}^{0}$

Step 5.2

Substitute and simplify.

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Step 5.2.1

Evaluate

$\frac{\arctan (\frac{x}{4})}{4}$ at

$0$ and at

$t$ .

$lim_{t \to - \infty} (\frac{\arctan (\frac{0}{4})}{4}) - \frac{\arctan (\frac{t}{4})}{4}$

Step 5.2.2

Cancel the common factor of

$0$ and

$4$ .

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Step 5.2.2.1

Factor

$4$ out of

$0$ .

$lim_{t \to - \infty} \frac{\arctan (\frac{4 (0)}{4})}{4} - \frac{\arctan (\frac{t}{4})}{4}$

Step 5.2.2.2

Cancel the common factors.

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Step 5.2.2.2.1

Factor

$4$ out of

$4$ .

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

Step 5.2.2.2.2

Cancel the common factor.

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

Step 5.2.2.2.3

Rewrite the expression.

$lim_{t \to - \infty} \frac{\arctan (\frac{0}{1})}{4} - \frac{\arctan (\frac{t}{4})}{4}$

Step 5.2.2.2.4

Divide

$0$ by

$1$ .

$lim_{t \to - \infty} \frac{\arctan (0)}{4} - \frac{\arctan (\frac{t}{4})}{4}$

Step 6

Evaluate the limit.

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Step 6.1

Combine fractions using a common denominator.

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Step 6.1.1

Combine the numerators over the common denominator.

$lim_{t \to - \infty} \frac{\arctan (0) - \arctan (\frac{t}{4})}{4}$

Step 6.1.2

Simplify the numerator.

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Step 6.1.2.1

The exact value of

$\arctan (0)$ is

$0$ .

$lim_{t \to - \infty} \frac{0 - \arctan (\frac{t}{4})}{4}$

Step 6.1.2.2

Subtract

$\arctan (\frac{t}{4})$ from

$0$ .

$lim_{t \to - \infty} \frac{- \arctan (\frac{t}{4})}{4}$

Step 6.1.3

Move the negative in front of the fraction.

$lim_{t \to - \infty} - \frac{\arctan (\frac{t}{4})}{4}$

Step 6.2

Move the term

$- 1$ outside of the limit because it is constant with respect to

$t$ .

$- lim_{t \to - \infty} \frac{\arctan (\frac{t}{4})}{4}$

Step 6.3

Move the term

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

$t$ .

$- \frac{1}{4} lim_{t \to - \infty} \arctan (\frac{t}{4})$

Step 6.4

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

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

Step 6.5

Substitute

$t$ for

$\frac{t}{4}$ and let

$t$ approach

$- \infty$ since

$lim_{t \to - \infty} \frac{t}{4} = - \infty$ .

$- \frac{1}{4} lim_{t \to - \infty} \arctan (t)$

Step 6.6

The limit as

$t$ approaches

$- \infty$ is

$- \frac{π}{2}$ .

$- \frac{1}{4} \cdot - 1 \frac{π}{2}$

Step 6.7

Simplify the answer.

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Step 6.7.1

Multiply

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

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Step 6.7.1.1

Multiply

$- 1$ by

$- 1$ .

$1 (\frac{1}{4}) \frac{π}{2}$

Step 6.7.1.2

Multiply

$\frac{1}{4}$ by

$1$ .

$\frac{1}{4} \cdot \frac{π}{2}$

Step 6.7.2

Multiply

$\frac{1}{4} \cdot \frac{π}{2}$ .

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Step 6.7.2.1

Multiply

$\frac{1}{4}$ by

$\frac{π}{2}$ .

$\frac{π}{4 \cdot 2}$

Step 6.7.2.2

Multiply

$4$ by

$2$ .

$\frac{π}{8}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{π}{8}$

Decimal Form:

$0.39269908 \dots$