Calculus Examples

$f (x) = e^{x}$ , $[- 2, 2]$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$e^{x} = 0$

Step 1.2

Solve $e^{x} = 0$ for $x$ .

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Step 1.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{x}) = \ln (0)$

Step 1.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 1.2.3

There is no solution for $e^{x} = 0$

No solution

Step 2

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{- 2}^{2} e^{x} d x - \int_{- 2}^{2} 0 d x$

Step 3

Integrate to find the area between $- 2$ and $2$ .

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Step 3.1

Combine the integrals into a single integral.

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

Step 3.2

Subtract $0$ from $e^{x}$ .

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

Step 3.3

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

$e^{x}]_{- 2}^{2}$

Step 3.4

Evaluate $e^{x}$ at $2$ and at $- 2$ .

$e^{2} - e^{- 2}$

Step 4

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

$e^{2} - \frac{1}{e^{2}}$

Step 5