Calculus Examples

$\int_{0}^{52} 0.005 t^{2} - 0.3 t - 1.5 d x$

Step 1

Apply the constant rule.

$(0.005 t^{2} - 0.3 t - 1.5) x]_{0}^{52}$

Step 2

Substitute and simplify.

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Step 2.1

Evaluate $(0.005 t^{2} - 0.3 t - 1.5) x$ at $52$ and at $0$ .

$((0.005 t^{2} - 0.3 t - 1.5) \cdot 52) - (0.005 t^{2} - 0.3 t - 1.5) \cdot 0$

Step 2.2

Simplify.

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Step 2.2.1

Move $52$ to the left of $0.005 t^{2} - 0.3 t - 1.5$ .

$52 \cdot (0.005 t^{2} - 0.3 t - 1.5) - (0.005 t^{2} - 0.3 t - 1.5) \cdot 0$

Step 2.2.2

Multiply $0$ by $- 1$ .

$52 (0.005 t^{2} - 0.3 t - 1.5) + 0 (0.005 t^{2} - 0.3 t - 1.5)$

Step 2.2.3

Multiply $0$ by $0.005 t^{2} - 0.3 t - 1.5$ .

$52 (0.005 t^{2} - 0.3 t - 1.5) + 0$

Step 2.2.4

Add $52 (0.005 t^{2} - 0.3 t - 1.5)$ and $0$ .

$52 (0.005 t^{2} - 0.3 t - 1.5)$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$52 (0.005 t^{2}) + 52 (- 0.3 t) + 52 \cdot - 1.5$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $0.005$ by $52$ .

$0.26 t^{2} + 52 (- 0.3 t) + 52 \cdot - 1.5$

Step 3.2.2

Multiply $- 0.3$ by $52$ .

$0.26 t^{2} - 15.6 t + 52 \cdot - 1.5$

Step 3.2.3

Multiply $52$ by $- 1.5$ .

$0.26 t^{2} - 15.6 t - 78$

Step 4