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

Question

Emilia Aguilar · Accepted Answer

#int_(-1)^(1)(root3(t)-2 )dt# Rewritten:
#=int_(-1)^(1)(t^(1/3)-2)dt# #=[3/4t^(4/3)-2t]_(-1)^1# #=(3/4(1)^(4/3)-2(1))-(3/4(-1)^(4/3)-2(-1))# #=color(red)(3/4)-2color(red)(-3/4)-2# #=-4#

Ezekiel Alexander · Answer

undefined 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 $.