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 .

Let's tackle each of the integration problems step by step. --- ### **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. --- ### **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:** \[ \int_{0}^{\pi/3} 4 \cos x \, \mathrm{d}x = 2\sqrt{3} \] --- ### **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:** \[ a \approx 2.36 \] --- **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. \[ a \approx 2.36 \]