Evaluate the integral.
-$\int t \sqrt { 3 } - 1 d t$
A) 1
B) $\frac { t \sqrt { 3 } } { \sqrt { 3 } } + C$
C) $\frac { t ^ { \sqrt { 3 } - 1 } } { \ln t } + C$
D) $\frac { t \sqrt { 3 } - 2 } { \sqrt { 3 } - 2 } + C$

Question

Quizplus · Accepted Answer

Use the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$.
In this problem, $n = \frac{1}{2}$, so we can write:
$$\int t \sqrt { 3 } - 1 d t = \int \sqrt{3} t - t dt = \frac{t^{2}}{2} \sqrt { 3 } - \frac{t^{2}}{2} + C = \frac { t \sqrt { 3 } } { \sqrt { 3 } } - \frac{ t } { 2 } + C = \frac { t \sqrt { 3 } } { 2 } - \frac { t } { 2 } + C = \frac { t \sqrt { 3 } - 1 } { 2 } + C$$
Therefore, the answer is B.

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