(a) Find $g ( t )$ so that $\int g ( t ) d t = \frac { t } { 2 } \sqrt { 4 - t ^ { 2 } } + 2 \cdot \sin ^ { - 1 } \frac { t } { 2 }$ (b) Evaluate $\int _ { 0 } ^ { 2 } g ( t ) d t$ using the Evaluation Theorem.(c) Evaluate $\int _ { 0 } ^ { 2 } g ( t ) d t$ by the area interpretation of the integral. Compare your answer with part (b) above.(d) Evaluate $\int _ { 0 } ^ { \sqrt { 3 } } g ( t ) d t .$

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The Answer of (a) Find $g ( t )$ so that $\int g ( t ) d t = \frac { t } { 2 } \sqrt { 4 - t ^ { 2 } } + 2 \cdot \sin ^ { - 1 } \frac { t } { 2 }$ (b) Evaluate $\int _ { 0 } ^ { 2 } g ( t ) d t$ using the Evaluation Theorem.(c) Evaluate $\int _ { 0 } ^ { 2 } g ( t ) d t$ by the area interpretation of the integral. Compare your answer with part (b) above.(d) Evaluate $\int _ { 0 } ^ { \sqrt { 3 } } g ( t ) d t .$

(A) Find g(t)g ( t )g(t) So That ∫g(t)dt=t24−t2+2⋅sin⁡−1t2\int g ( t ) d t = \frac { t } { 2 } \sqrt { 4 - t ^ { 2 } } + 2 \cdot \sin ^ { - 1 } \frac { t } { 2 }∫g(t)dt=2t​4−t2​+2⋅sin−12t​

(A) Find $g ( t )$ So That $\int g ( t ) d t = \frac { t } { 2 } \sqrt { 4 - t ^ { 2 } } + 2 \cdot \sin ^ { - 1 } \frac { t } { 2 }$