# You notice a bug flying around the room. The distance s(t) (in feet) that the bug has traveled...

## Question:

You notice a bug flying around the room. The distance s(t) (in feet) that the bug has traveled after t seconds is given by

{eq}s\left ( t \right ) = \int_{0}^{3t^2}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx {/eq}

a. Find the speed {eq}\frac{\mathrm{d} s}{\mathrm{d} t} {/eq} of the bug at time {eq}t = \frac{1}{2} {/eq} seconds.

b. After t = 1 second, the bug has traveled a distance.

{eq}s\left ( 1 \right ) = \int_{0}^{3}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx {/eq}

## Fundamental Theorem of Calculus - Part I

Part I of the Fundamental Theorem of Calculus states that

{eq}\displaystyle \frac{d}{dx}\left(\int_a^x f(t)\ dt \right)=f(x), \text{ where } a \text{ is constant}. {/eq}

Or the Chain rule form {eq}\displaystyle \frac{d}{dx}\left(\int_a^{u(x)} f(t)\ dt\right) =f(u(x))\cdot u'(x), \text{ where } '=\frac{d}{dx}. {/eq}

It is important to have a constant as a lower limit of integration and as the upper limit of integration,

to have a function of {eq}x {/eq}, the variable with respect to which the derivative is taken.

If not, we need to use the properties of the definite integrals to bring it to the required form.

## Answer and Explanation:

a. The speed of the bug whose position function is given as {eq}\displaystyle s\left ( t \right ) = \int_{0}^{3t^2}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx {/eq}

is obtained by taken the derivative with respect to {eq}\displaystyle t, {/eq} as below.

{eq}\displaystyle \begin{align} \text{speed }&=s'(t)=\frac{d}{dt}\left(\int_{0}^{3t^2}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx\right)\\ &= \sqrt{2(3t^2)^2 + cos^2\left ( \pi \cdot 3t^2 \right )}\cdot \frac{d}{dt}\left(3t^2\right), &\left[\text{ using the Fundamental theorem of calculus part I}\right]\\ &= 6t\sqrt{18t^4 + \cos^2\left (3 \pi t^2 \right )}\\ \implies s'\left(\frac{1}{2}\right)&=3\sqrt{18\cdot \frac{1}{4} + \cos^2\left (3 \pi \frac{1}{4} \right )}\\ &=3\sqrt{\frac{9}{2} + \frac{1}{2} }\\ &=\boxed{3\sqrt{5}}. \end{align} {/eq}

b. The distance traveled in the first one second is {eq}\displaystyle s\left ( 1 \right ) = \int_{0}^{3}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx\boxed{\approx 6.88} {/eq} using numerical integration, Riemann sum, because otherwise is challenging to solve it analytically.