Question

Let f : [1, 2] → \(\R\) be the function defined by
\(f(t)=\int^t_1\sqrt{x^2e^{x^2}-1}\ dx.\)
Then the arc length of the graph of f over the interval [1, 2] equals

• A. \(e^2-\sqrt{e}\)
• B. \(e-\sqrt{e}\)
• C. e² - e
• D. e² - 1

Answer 1

The Correct Option is A

To find the arc length of the graph of the function \( f(t) = \int_{1}^{t} \sqrt{x^2 e^{x^2} - 1} \, dx \) over the interval [1, 2], we need to apply the formula for the arc length of a function defined by an integral.

The general formula for the arc length \( L \) of a function \( f(t) \) over an interval \([a, b]\) is given by:

\(L = \int_{a}^{b} \sqrt{1 + \left(\frac{df}{dt}\right)^2} \, dt\)

Here, we first need to calculate \( \frac{df}{dt} \) for the function \( f(t) \).

Given that \( f(t) = \int_{1}^{t} \sqrt{x^2 e^{x^2} - 1} \, dx \), by the Fundamental Theorem of Calculus, the derivative of \( f(t) \) is:

\(\frac{df}{dt} = \sqrt{t^2 e^{t^2} - 1}\)

Now, substitute \( \frac{df}{dt} \) into the arc length formula:

\(L = \int_{1}^{2} \sqrt{1 + \left(\sqrt{t^2 e^{t^2} - 1}\right)^2} \, dt\)

Simplify the expression inside the square root:

\(\sqrt{1 + t^2 e^{t^2} - 1} = \sqrt{t^2 e^{t^2}}\)

Thus, the integral simplifies to:

\(L = \int_{1}^{2} t e^{t^2/2} \, dt\)

This integral can be solved using substitution. Let \( u = t^2/2 \), then \( du = t \, dt \), and the limits change accordingly from \( t = 1 \) to \( t = 2 \).

When \( t = 1 \), \( u = 1^2/2 = 1/2 \)

When \( t = 2 \), \( u = 2^2/2 = 2 \)

The integral in terms of \( u \) becomes:

\(\int_{1/2}^{2} e^u \, du = e^u \bigg|_{1/2}^{2}\)

Calculating this, we get:

\(e^2 - e^{1/2} = e^2 - \sqrt{e}\)

Therefore, the arc length of the graph of the function \( f(t) \) over the interval [1, 2] is \(e^2 - \sqrt{e}\).