FORM GR1268 Question 14

Question:
Suppose \(g\) is a continuous real-valued function such that \(\displaystyle3x^5+96=\int_c^x g(t)dt\) for each \(x\in\mathbb R\), where \(c\) is a constant. What is the value of \(c\)?

(A) \(-96\)     (B) \(-2\)     (C) \(4\)     (D) \(15\)     (E) \(32\)

Answer:
(B)

Answer Key:

Method 1
Because the equation is true for any real number, it is true for \(x=c\).

Then \(\displaystyle 3c^5+96=\int_c^c g(t)dt=0\).

Thus, \(3c^5=-96\), so \(c^5=-32\), so \(c=(-32)^{1/5}=-2\).

Method 2
Differentiate both sides with respect to \(x\) to obtain \(15x^4=g(x)\).
(See note below if you are not sure about this step.)

Then
\[\begin{align}
3x^5+96&=\int_c^x g(t)dt\\
&= \int_c^x15t^4dt\\
&=\left.3t^5\right|_c^x\\
&=3x^5-3c^5.
\end{align}\]
Thus, \(-3c^5=96\), so \(c^5=-32\), so \(c=(-32)^{1/5}=-2\).

Note on differentiating integral:
Suppose the antiderivative of \(g(t)\) is \(G(t)\).

Then \(\displaystyle\int_c^xg(t)dt=\left.G(t)\right|_c^x=G(x)-G(c)\).

Since \(G(c)\) is a constant, \(\displaystyle\frac{d}{dx}\int_c^xg(t)dt=\frac{d}{dx}[G(x)-G(c)]=g(x)\).