Question

Suppose that \(\int _ { - 3 } ^ { 0 } g ( t ) d t = \sqrt { 2 } .\) Find each integral. \(\begin{array} { l l } { \text { (a) } \int _ { 0 } ^ { - 3 } g ( t ) d t } & { \text { (b) } \int _ { - 3 } ^ { 0 } g ( u ) d u } \\ { \text { (c) } \int _ { - 3 } ^ { 0 } [ - g ( x ) ] d x } & { \text { (d) } \int _ { - 3 } ^ { 0 } \frac { g ( r ) } { \sqrt { 2 } } d r } \end{array}\)

Short answer

(a) -\( \sqrt {2} \) (b) \( \sqrt {2} \) (c) -\( \sqrt {2} \) (d) 1.

Step-by-step solution

Change of Limits

The change of limits from -3 to 0, to 0 to -3 changes the sign of the integral. Therefore, \[ \int _ { 0 } ^ { - 3 } g ( t ) d t = -\sqrt { 2 } . \]

Variable of Integration

The variable of integration can be changed without changing the result if the limits remain the same. Therefore, \[ \int _ { - 3 } ^ { 0 } g ( u ) d u = \sqrt {2} . \]

Negative of the Function

Integrating the negative of the function is equivalent to changing the sign of the integral. As a result, \[ \int _ { - 3 } ^ { 0 } [ - g ( x ) ] d x = -\sqrt {2} . \]

Scaling of the Function

Multiplying a function by a constant inside the integral scales the integral. Consequently, \[ \int _ { - 3 } ^ { 0 } \frac { g ( r ) } { \sqrt { 2 } } d r = 1 . \]