Calculus Examples

$\int_{0}^{10} - 0.006 t^{2} + 0.2 t d x$

Step 1

Split the single integral into multiple integrals.

$\int_{0}^{10} - 0.006 t^{2} d x + \int_{0}^{10} 0.2 t d x$

Step 2

Apply the constant rule.

$- 0.006 t^{2} x]_{0}^{10} + \int_{0}^{10} 0.2 t d x$

Step 3

Apply the constant rule.

$- 0.006 t^{2} x]_{0}^{10} + 0.2 t x]_{0}^{10}$

Step 4

Simplify the answer.

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Step 4.1

Combine $- 0.006 t^{2} x]_{0}^{10}$ and $0.2 t x]_{0}^{10}$ .

$- 0.006 t^{2} x + 0.2 t x]_{0}^{10}$

Step 4.2

Substitute and simplify.

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Step 4.2.1

Evaluate $- 0.006 t^{2} x + 0.2 t x$ at $10$ and at $0$ .

$(- 0.006 t^{2} \cdot 10 + 0.2 t \cdot 10) - (- 0.006 t^{2} \cdot 0 + 0.2 t \cdot 0)$

Step 4.2.2

Simplify.

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Step 4.2.2.1

Multiply $10$ by $- 0.006$ .

$- 0.06 t^{2} + 0.2 t \cdot 10 - (- 0.006 t^{2} \cdot 0 + 0.2 t \cdot 0)$

Step 4.2.2.2

Multiply $10$ by $0.2$ .

$- 0.06 t^{2} + 2 t - (- 0.006 t^{2} \cdot 0 + 0.2 t \cdot 0)$

Step 4.2.2.3

Multiply $0$ by $- 0.006$ .

$- 0.06 t^{2} + 2 t - (0 t^{2} + 0.2 t \cdot 0)$

Step 4.2.2.4

Multiply $0$ by $t^{2}$ .

$- 0.06 t^{2} + 2 t - (0 + 0.2 t \cdot 0)$

Step 4.2.2.5

Multiply $0$ by $0.2$ .

$- 0.06 t^{2} + 2 t - (0 + 0 t)$

Step 4.2.2.6

Multiply $0$ by $t$ .

$- 0.06 t^{2} + 2 t - (0 + 0)$

Step 4.2.2.7

Add $0$ and $0$ .

$- 0.06 t^{2} + 2 t - 0$

Step 4.2.2.8

Multiply $- 1$ by $0$ .

$- 0.06 t^{2} + 2 t + 0$

Step 4.2.2.9

Add $- 0.06 t^{2} + 2 t$ and $0$ .

$- 0.06 t^{2} + 2 t$

Step 5