Fluency exercise

# Integral chasing II Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Solution

Can you find the missing positive numbers $a$ to $d$?

We will work through the integrals in order, but they can be tackled in a variety of ways.

1. $\displaystyle{\int_1^a \dfrac{1}{3x} \,dx = \ln 2}$
\begin{align*} \int_1^a \dfrac{1}{3x} \,dx &= \left[\frac{1}{3}\ln(3x)\right]_{1}^{a} \\ &= \frac{1}{3}\big(\ln3a - \ln3\big) \\ &= \frac{1}{3}\ln\left(\frac{3a}{3}\right) \\ &= \ln{a^{\frac{1}{3}}} \end{align*}

So $\ln{a^{\frac{1}{3}}} = \ln2$ which gives us $a = 8$.

We have treated this integral as $\displaystyle{\int_1^a \dfrac{1}{(3x)} \,dx}$, but what happens if we think of it as $\displaystyle{\dfrac{1}{3}\int_1^a \dfrac{1}{x} \,dx}$? Do both approaches give the same answer?

1. $\displaystyle{\int_0^3 ax e^{2x} \,dx = be^6 + 2}$

What method will we use for integrating this? We have a product of two different functions so it looks like integration by parts will be useful. $\int uv' \, dx = uv - \int u'v \, dx$ Which function will we integrate and which will we differentiate?

Since we know that $a = 8$, we have two functions, $8x$ and $e^{2x}$. We can differentiate both these functions, but $8x$ will become a constant so let’s take $u = 8x$ and $v' = e^{2x}$. That gives us $u'=8$ and $v = \frac{1}{2}e^{2x}$. Therefore the integral can be written as:

\begin{align*} \int_0^3 8x e^{2x} \,dx &= \left[4xe^{2x}\right]_0^3 - \int_0^3 4e^{2x} \, dx \\ &= \left[4xe^{2x} - 2e^{2x} \right]_0^3 \\ &= (12e^6 - 2e^6) - (0 - 2e^0) \\ &= 10e^6 + 2. \end{align*}

Comparing the answers we have $10e^6+2 = be^6 + 2$, so $b = 10$.

1. $\displaystyle{\int_{1}^{3} \dfrac{cx}{\sqrt{25-x^2}} \,dx = 4\sqrt{6} - 4c}$

The function in the numerator is a multiple of the derivative of the function in the square root, so it seems reasonable to think that we should try integration by substitution.

Let $u = 25-x^2$, then $du = -2x \ dx$. Some manipulation is required to get the integral in the form we need.

$\int_{1}^{3} \dfrac{cx}{\sqrt{25-x^2}} \,dx = -\dfrac{c}{2} \int_{1}^{3} \dfrac{-2x}{\sqrt{25-x^2}} \, dx$

Now we can make the substitution. \begin{align*} -\dfrac{c}{2} \int_{1}^{3} \dfrac{-2x}{\sqrt{25-x^2}} \, dx &= -\dfrac{c}{2} \int_{u = 24}^{u=16} \dfrac{1}{\sqrt{u}} \, du \\ &= -\dfrac{c}{2} \left[2u^{\frac{1}{2}}\right]_{u=24}^{u=16} \\ &= -c \big(\sqrt{16} - \sqrt{24} \big) \\ &= 2c\sqrt{6} - 4c \end{align*}

If we compare this to the answer given we have $4\sqrt{6} - 4c = 2c\sqrt{6} - 4c$, so $c = 2$.

When using integration by substitution there is likely to be more than one possible substitution that could be used. What other substitutions could be used here?

1. $\displaystyle{\int_{\frac{\pi}{2}}^{\frac{b}{a}\pi} x^2\sin cx \,dx = \dfrac{\pi d - \pi^2 + c}{a}}$

As in the second integral we have the product of two unrelated functions so it makes sense to try integration by parts. Which way round should the choices for $u$ and $v'$ be?

We start by substituting in the values of $a, b$ and $c$.

$\int_{\frac{\pi}{2}}^{\frac{5}{4}\pi} x^2\sin 2x \,dx$

If $u = x^2$ and $v' = \sin2x$ then we have $u' = 2x$ and $v = -\frac{1}{2}\cos2x$.

\begin{align*} \int_{\frac{\pi}{2}}^{\frac{5}{4}\pi} x^2\sin 2x \,dx &= \left[-\frac{1}{2}x^2\cos2x\right]_{\frac{\pi}{2}}^{\frac{5}{4}\pi} - \int_{\frac{\pi}{2}}^{\frac{5}{4}\pi} -x\cos2x \, dx \end{align*}

Using integration by parts hasn’t reduced this to an integral that we can now solve but it has reduced the polynomial term from $x^2$ to $x$.

While we can’t solve this directly, it has got us one step closer. It is now similar to the second integral and we can use integration by parts for a second time where $u = x$ and $v'= \cos2x$, and $u'=1$ and $v = \frac{1}{2}\sin2x$.

\begin{align*} \left[-\frac{1}{2}x^2\cos2x\right]_{\frac{\pi}{2}}^{\frac{5}{4}\pi} + \int_{\frac{\pi}{2}}^{\frac{5}{4}\pi} x\cos2x \, dx &= \left[-\frac{1}{2}x^2\cos2x + \frac{1}{2}x\sin2x\right]_{\frac{\pi}{2}}^{\frac{5}{4}\pi} - \int_{\frac{\pi}{2}}^{\frac{5}{4}\pi} \frac{1}{2}\sin2x \, dx \\ &= \left[-\frac{1}{2}x^2\cos2x + \frac{1}{2}x\sin2x\ + \frac{1}{4}\cos2x\right]_{\frac{\pi}{2}}^{\frac{5}{4}\pi} \end{align*}

Before we substitute in the limits it is worth noticing that all our trigonometric functions have an argument of $2x$. When substituting $x =\frac{5\pi}{4}$ or $x = \frac{\pi}{2}$ into the trigonometric functions we will get the following values.

$\cos(\frac{5\pi}{2}) = 0$

$\sin(\frac{5\pi}{2}) = 1$

$\sin\pi = 0$

$\cos\pi = -1$

Therefore when we evaluate the expression we get

$\left(0+\frac{1}{2}\times\dfrac{5\pi}{4}\times1 + 0\right) - \left(-\dfrac{1}{2}\times(\dfrac{\pi}{2})^2\times-1 + 0 + \dfrac{1}{4}\times1\right) = \dfrac{5\pi}{8} - \dfrac{\pi^2}{8} + \dfrac{1}{4}.$

We can use the answer given in the question to check our values for $a, b$ and $c$ and to find the value for $d$.

\begin{align*} \dfrac{\pi d - \pi^2 + c}{a} &= \dfrac{5\pi}{8} - \dfrac{\pi^2}{8} + \dfrac{1}{4} \\ &= \dfrac{5\pi - \pi^2 +2}{8} \end{align*}

We can be confident that $d = 5$ since our values for $a$ and $c$ match what we have already found.