# How do you evaluate the definite integral #int root3(t)-2dt# from [-1,1]?

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To evaluate the definite integral ( \int_{-1}^{1} (\sqrt{3t} - 2) , dt ), you first need to find the antiderivative of ( \sqrt{3t} - 2 ) with respect to ( t ), and then evaluate it at the upper and lower limits of integration and subtract the results.

Step 1: Find the antiderivative of ( \sqrt{3t} - 2 ): [ \int (\sqrt{3t} - 2) , dt = \frac{2}{3} (3t)^{\frac{3}{2}} - 2t + C ]

Step 2: Evaluate the antiderivative at the upper and lower limits of integration: [ \left[ \frac{2}{3} (3t)^{\frac{3}{2}} - 2t \right]_{-1}^{1} ]

Step 3: Substitute the upper and lower limits into the antiderivative expression and subtract: [ = \left[ \frac{2}{3} (3(1))^{\frac{3}{2}} - 2(1) \right] - \left[ \frac{2}{3} (3(-1))^{\frac{3}{2}} - 2(-1) \right] ]

[ = \left[ \frac{2}{3} (3)^{\frac{3}{2}} - 2 \right] - \left[ \frac{2}{3} (-3)^{\frac{3}{2}} + 2 \right] ]

[ = \left[ \frac{2}{3} (3)^{\frac{3}{2}} - 2 \right] - \left[ -\frac{2}{3} (3)^{\frac{3}{2}} + 2 \right] ]

[ = \frac{4}{3} (3)^{\frac{3}{2}} - 4 ]

[ = \frac{4}{3} \cdot 3\sqrt{3} - 4 ]

[ = 4\sqrt{3} - 4 ]

Therefore, ( \int_{-1}^{1} (\sqrt{3t} - 2) , dt = 4\sqrt{3} - 4 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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