Integration Prep 1. Find $$ \int \frac{\sqrt{x}-3 x^{2}}{4 x^{3}} \mathrm{dx} $$ (accessible to students on the path to grade 3 or 4) [6 marks] 2. Do not use a calculator to answer this question. Find the exact value of $$ \int_{0}^{\pi / 3} 4 \cos x \mathrm{~d} x $$. (accessible to students on the path to grade 3 or 4 ) [5 marks] 3. The diagram shows the graph of $$ y=\mathrm{e}^{x}-x $$. Find the value of $$ a $$ such that the shaded area equals 6.8 .

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Let's tackle each of the integration problems step by step.

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### **1. Evaluate the Integral**

$$ \int \frac{\sqrt{x} - 3x^{2}}{4x^{3}} \, \mathrm{dx} $$

**Step 1: Simplify the Integrand**

First, rewrite the integrand by expressing $$\sqrt{x}$$ and simplifying the fractions:

$$
\frac{\sqrt{x} - 3x^{2}}{4x^{3}} = \frac{x^{1/2}}{4x^{3}} - \frac{3x^{2}}{4x^{3}} = \frac{x^{-5/2}}{4} - \frac{3x^{-1}}{4}
$$

**Step 2: Integrate Term by Term**

Now, integrate each term separately:

$$
\int \left( \frac{1}{4}x^{-5/2} - \frac{3}{4}x^{-1} \right) \, \mathrm{dx} = \frac{1}{4} \int x^{-5/2} \, \mathrm{dx} - \frac{3}{4} \int x^{-1} \, \mathrm{dx}
$$

Compute each integral:

1. **First Integral:**
   $$
   \frac{1}{4} \int x^{-5/2} \, \mathrm{dx} = \frac{1}{4} \cdot \left( \frac{x^{-3/2}}{-3/2} \right) = -\frac{1}{6} x^{-3/2}
   $$

2. **Second Integral:**
   $$
   -\frac{3}{4} \int x^{-1} \, \mathrm{dx} = -\frac{3}{4} \ln|x|
   $$

**Step 3: Combine the Results**

Putting it all together, the integral becomes:

$$
\int \frac{\sqrt{x} - 3x^{2}}{4x^{3}} \, \mathrm{dx} = -\frac{1}{6} \cdot \frac{1}{x^{3/2}} - \frac{3}{4} \ln|x| + C
$$

where $$C$$ is the constant of integration.

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### **2. Compute the Definite Integral**

$$ \int_{0}^{\pi/3} 4 \cos x \, \mathrm{d}x $$

**Step 1: Find the Antiderivative**

The antiderivative of $$4 \cos x$$ is:

$$
\int 4 \cos x \, \mathrm{d}x = 4 \sin x + C
$$

**Step 2: Evaluate from $$0$$ to $$\pi/3$$**

Apply the limits of integration:

$$
4 \sin \left( \frac{\pi}{3} \right) - 4 \sin(0) = 4 \left( \frac{\sqrt{3}}{2} \right) - 0 = 2\sqrt{3}
$$

**Final Answer:**

$$
\int_{0}^{\pi/3} 4 \cos x \, \mathrm{d}x = 2\sqrt{3}
$$

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### **3. Determine the Value of $$a$$ for a Given Area**

**Given:**
The area under the curve $$ y = e^{x} - x $$ from $$0$$ to $$a$$ is $$6.8$$.

**Set Up the Integral:**

$$
\int_{0}^{a} (e^{x} - x) \, \mathrm{d}x = 6.8
$$

**Compute the Integral:**

$$
\int (e^{x} - x) \, \mathrm{d}x = e^{x} - \frac{1}{2}x^{2} + C
$$

Apply the limits from $$0$$ to $$a$$:

$$
\left( e^{a} - \frac{1}{2}a^{2} \right) - \left( e^{0} - 0 \right) = 6.8
$$
$$
e^{a} - \frac{1}{2}a^{2} - 1 = 6.8
$$
$$
e^{a} - \frac{1}{2}a^{2} = 7.8
$$

**Solve for $$a$$:**

This equation is transcendental and generally requires numerical methods to solve. However, we can approximate:

1. **Try $$a = 2$$:**
   $$
   e^{2} - \frac{1}{2}(2)^2 \approx 7.389 - 2 = 5.389 \quad (\text{Too low})
   $$

2. **Try $$a = 2.3$$:**
   $$
   e^{2.3} - \frac{1}{2}(2.3)^2 \approx 9.974 - 2.645 = 7.329 \quad (\text{Close to 7.8})
   $$

3. **Try $$a = 2.36$$:**
   $$
   e^{2.36} - \frac{1}{2}(2.36)^2 \approx 10.61 - 2.785 = 7.825 \quad (\text{Very close to 7.8})
   $$

**Final Approximation:**

$$
a \approx 2.36
$$

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**Summary of Answers:**

1. $$
   \int \frac{\sqrt{x}-3x^{2}}{4x^{3}} \, \mathrm{dx} = -\frac{1}{6x^{3/2}} - \frac{3}{4} \ln|x| + C
   $$

2. $$
   \int_{0}^{\pi/3} 4 \cos x \, \mathrm{d}x = 2\sqrt{3}
   $$

3. $$
   a \approx 2.36
   $$
1. The integral of $$ \frac{\sqrt{x} - 3x^{2}}{4x^{3}} $$ with respect to $$ x $$ is $$ -\frac{1}{6x^{3/2}} - \frac{3}{4} \ln|x| + C $$. 2. The exact value of $$ \int_{0}^{\pi/3} 4 \cos x \, \mathrm{d}x $$ is $$ 2\sqrt{3} $$. 3. The value of $$ a $$ such that the shaded area under $$ y = e^{x} - x $$ from $$0$$ to $$a$$ equals $$6.8$$ is approximately $$2.36$$.

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