TMUA practice: Integration Power Rule, problem 1

Integrate $g'$ to get $g$ up to a constant.

$g(x) \;=\; \int (4 x^{3} + 6 x - 1)\, dx \;=\; x^{4} + 3 x^{2} - x + C.$

Apply the condition $g(1) = 5$ to find $C$:

$g(1) \;=\; 1 + 3 - 1 + C \;=\; 3 + C \;=\; 5 \;\Longrightarrow\; C = 2.$

So $g(x) = x^{4} + 3 x^{2} - x + 2$.

Evaluate at $x = 2$:

$g(2) \;=\; 16 + 12 - 2 + 2 \;=\; 28.$

This is option D.

Why the other options fail.

• A, $16$: equals $x^{4}$ at $x = 2$ alone; ignored the rest of $g$.
• B, $22$: forgot the constant $C$ (used $C = 0$ instead of $C = 2$, then $g(2) = 16 + 12 - 2 = 26$). Or arithmetic slip $26 - 4 = 22$.
• C, $26$: as above with $C = 0$.
• E, $30$: $g(2) + 2$, double-counting the constant.

The lesson: $g'$ pins down $g$ only up to a constant. Use the boundary condition $g(x_{0}) = y_{0}$ to nail down $C$ before evaluating elsewhere.