Calculus Examples

$f (t) = \frac{2 t - 4 + 4 \sqrt{t}}{\sqrt{t}}$

Step 1

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

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

Step 2

Set up the integral to solve.

$F (t) = \int \frac{2 t - 4 + 4 \sqrt{t}}{\sqrt{t}} d t$

Step 3

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t}$ as $t^{\frac{1}{2}}$ .

$\int \frac{2 t - 4 + 4 t^{\frac{1}{2}}}{\sqrt{t}} d t$

Step 4

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t}$ as $t^{\frac{1}{2}}$ .

$\int \frac{2 t - 4 + 4 t^{\frac{1}{2}}}{t^{\frac{1}{2}}} d t$

Step 5

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

$\int (2 t - 4 + 4 t^{\frac{1}{2}}) {(t^{\frac{1}{2}})}^{- 1} d t$

Step 6

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

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Step 6.1

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

$\int (2 t - 4 + 4 t^{\frac{1}{2}}) t^{\frac{1}{2} \cdot - 1} d t$

Step 6.2

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

$\int (2 t - 4 + 4 t^{\frac{1}{2}}) t^{\frac{- 1}{2}} d t$

Step 6.3

Move the negative in front of the fraction.

$\int (2 t - 4 + 4 t^{\frac{1}{2}}) t^{- \frac{1}{2}} d t$

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

$\int (2 t - 4) t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.2

Apply the distributive property.

$\int 2 t \cdot t^{- \frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.3

Raise $t$ to the power of $1$ .

$\int 2 (t^{1} t^{- \frac{1}{2}}) - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 2 t^{1 - \frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.5

Write $1$ as a fraction with a common denominator.

$\int 2 t^{\frac{2}{2} - \frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.6

Combine the numerators over the common denominator.

$\int 2 t^{\frac{2 - 1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.7

Subtract $1$ from $2$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2}} t^{- \frac{1}{2}} d t$

Step 7.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1}{2} - \frac{1}{2}} d t$

Step 7.9

Combine the numerators over the common denominator.

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{1 - 1}{2}} d t$

Step 7.10

Subtract $1$ from $1$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{0}{2}} d t$

Step 7.11

Cancel the common factor of $0$ and $2$ .

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Step 7.11.1

Factor $2$ out of $0$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{2 (0)}{2}} d t$

Step 7.11.2

Cancel the common factors.

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Step 7.11.2.1

Factor $2$ out of $2$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{2 \cdot 0}{2 \cdot 1}} d t$

Step 7.11.2.2

Cancel the common factor.

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{2 \cdot 0}{2 \cdot 1}} d t$

Step 7.11.2.3

Rewrite the expression.

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{\frac{0}{1}} d t$

Step 7.11.2.4

Divide $0$ by $1$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 t^{0} d t$

Step 7.12

Anything raised to $0$ is $1$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 \cdot 1 d t$

Step 7.13

Multiply $4$ by $1$ .

$\int 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}} + 4 d t$

Step 8

Split the single integral into multiple integrals.

$\int 2 t^{\frac{1}{2}} d t + \int - 4 t^{- \frac{1}{2}} d t + \int 4 d t$

Step 9

Since $2$ is constant with respect to $t$ , move $2$ out of the integral.

$2 \int t^{\frac{1}{2}} d t + \int - 4 t^{- \frac{1}{2}} d t + \int 4 d t$

Step 10

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

$2 (\frac{2}{3} t^{\frac{3}{2}} + C) + \int - 4 t^{- \frac{1}{2}} d t + \int 4 d t$

Step 11

Since $- 4$ is constant with respect to $t$ , move $- 4$ out of the integral.

$2 (\frac{2}{3} t^{\frac{3}{2}} + C) - 4 \int t^{- \frac{1}{2}} d t + \int 4 d t$

Step 12

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

$2 (\frac{2}{3} t^{\frac{3}{2}} + C) - 4 (2 t^{\frac{1}{2}} + C) + \int 4 d t$

Step 13

Apply the constant rule.

$2 (\frac{2}{3} t^{\frac{3}{2}} + C) - 4 (2 t^{\frac{1}{2}} + C) + 4 t + C$

Step 14

Simplify.

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Step 14.1

Combine $\frac{2}{3}$ and $t^{\frac{3}{2}}$ .

$2 (\frac{2 t^{\frac{3}{2}}}{3} + C) - 4 (2 t^{\frac{1}{2}} + C) + 4 t + C$

Step 14.2

Simplify.

$\frac{4 t^{\frac{3}{2}}}{3} - 8 t^{\frac{1}{2}} + 4 t + C$

Step 14.3

Reorder terms.

$\frac{4}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + 4 t + C$

Step 15

The answer is the antiderivative of the function $f (t) = \frac{2 t - 4 + 4 \sqrt{t}}{\sqrt{t}}$ .

$F (t) =$ $\frac{4}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + 4 t + C$