Calculus Examples

$\sin (2 - x)$

Write $\sin (2 - x)$ as a function.

$f (x) = \sin (2 - x)$

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Set up the integral to solve.

$F (x) = \int \sin (2 - x) d x$

Let $u = 2 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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

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Differentiate $2 - x$ .

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

Differentiate.

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

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

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

$0 + \frac{d}{d x} [- x]$

Evaluate $\frac{d}{d x} [- x]$ .

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$0 - \frac{d}{d x} [x]$

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

$0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$0 - 1$

Subtract $1$ from $0$ .

$- 1$

Rewrite the problem using $u$ and $d u$ .

$\int - \sin (u) d u$

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

$- \int \sin (u) d u$

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

$- (- \cos (u) + C)$

Simplify.

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Simplify.

$- - \cos (u) + C$

Simplify.

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Multiply $- 1$ by $- 1$ .

$1 \cos (u) + C$

Multiply $\cos (u)$ by $1$ .

$\cos (u) + C$

Replace all occurrences of $u$ with $2 - x$ .

$\cos (2 - x) + C$

The answer is the antiderivative of the function $f (x) = \sin (2 - x)$ .

$F (x) =$ $\cos (2 - x) + C$