Calculus Examples

$\frac{d P}{d t} = - 5 t e^{- 0.2 t}$

Step 1

Rewrite the equation.

$d P = - 5 t e^{- 0.2 t} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d P = \int - 5 t e^{- 0.2 t} d t$

Step 2.2

Apply the constant rule.

$P + C_{1} = \int - 5 t e^{- 0.2 t} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $- 5$ is constant with respect to $t$ , move $- 5$ out of the integral.

$P + C_{1} = - 5 \int t e^{- 0.2 t} d t$

Step 2.3.2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = t$ and $d v = e^{- 0.2 t}$ .

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) - \int - 5 e^{- 0.2 t} d t)$

Step 2.3.3

Since $- 5$ is constant with respect to $t$ , move $- 5$ out of the integral.

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) - (- 5 \int e^{- 0.2 t} d t))$

Step 2.3.4

Multiply $- 5$ by $- 1$ .

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) + 5 \int e^{- 0.2 t} d t)$

Step 2.3.5

Let $u = - 0.2 t$ . Then $d u = - 0.2 d t$ , so $\frac{1}{- 0.2} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.5.1

Let $u = - 0.2 t$ . Find $\frac{d u}{d t}$ .

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Step 2.3.5.1.1

Differentiate $- 0.2 t$ .

$\frac{d}{d t} [- 0.2 t]$

Step 2.3.5.1.2

Since $- 0.2$ is constant with respect to $t$ , the derivative of $- 0.2 t$ with respect to $t$ is $- 0.2 \frac{d}{d t} [t]$ .

$- 0.2 \frac{d}{d t} [t]$

Step 2.3.5.1.3

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

$- 0.2 \cdot 1$

Step 2.3.5.1.4

Multiply $- 0.2$ by $1$ .

$- 0.2$

Step 2.3.5.2

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

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) + 5 \int e^{u} \frac{1}{- 0.2} d u)$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Move the negative in front of the fraction.

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) + 5 \int e^{u} (- \frac{1}{0.2}) d u)$

Step 2.3.6.2

Combine $e^{u}$ and $\frac{1}{0.2}$ .

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) + 5 \int - \frac{e^{u}}{0.2} d u)$

Step 2.3.7

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

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) + 5 (- \int \frac{e^{u}}{0.2} d u))$

Step 2.3.8

Multiply $- 1$ by $5$ .

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) - 5 \int \frac{e^{u}}{0.2} d u)$

Step 2.3.9

Since $\frac{1}{0.2}$ is constant with respect to $u$ , move $\frac{1}{0.2}$ out of the integral.

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) - 5 (\frac{1}{0.2} \int e^{u} d u))$

Step 2.3.10

Simplify.

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Step 2.3.10.1

Combine $\frac{1}{0.2}$ and $- 5$ .

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) + \frac{- 5}{0.2} \int e^{u} d u)$

Step 2.3.10.2

Move the negative in front of the fraction.

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) - \frac{5}{0.2} \int e^{u} d u)$

Step 2.3.11

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

$P + C_{1} = - 5 (t (- 5 e^{- 0.2 t}) - \frac{5}{0.2} (e^{u} + C_{2}))$

Step 2.3.12

Rewrite $- 5 (t (- 5 e^{- 0.2 t}) - \frac{5}{0.2} (e^{u} + C_{2}))$ as $- 5 (- 5 t e^{- 0.2 t} - 25 e^{u}) + C_{2}$ .

$P + C_{1} = - 5 (- 5 t e^{- 0.2 t} - 25 e^{u}) + C_{2}$

Step 2.3.13

Replace all occurrences of $u$ with $- 0.2 t$ .

$P + C_{1} = - 5 (- 5 t e^{- 0.2 t} - 25 e^{- 0.2 t}) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$P = - 5 (- 5 t e^{- 0.2 t} - 25 e^{- 0.2 t}) + K$