IB Mathematics SL 5.5 integration as anti-differentiation AI SL Paper 2 - Exam Style Questions - New Syllabus

(a)
Using a GDC to find the intersection of \(y = 1.5^x\) and \(y = 6 – 3/x\):
The x-coordinates are 0.638 (0.637623…) and 4.10 (4.09811…).

(b)
(i) The shaded region is under \(f(x)\) from \(x=0.638\) to \(x=4.10\).
Integral: \(\int_{0.638}^{4.10} 1.5^x \, dx\) (or \(\int_{0.638}^{4.10} f(x) \, dx\)).
(ii) Calculating the definite integral:
Area = 9.80 (9.81 or 9.80788…).
(iii) The region enclosed between the curves is \(\int_{0.638}^{4.10} (g(x) – f(x)) \, dx\).
Using GDC: Area = 5.38 (5.38290…).

(c)
Set the derivatives equal: \(f'(x) = g'(x)\).
\(f'(x) = 1.5^x \ln(1.5)\).
\(g'(x) = 3x^{-2} = \frac{3}{x^2}\).
Solve \(1.5^x \ln(1.5) = \frac{3}{x^2}\).
\(k = \mathbf{1.86}\) (1.86406…).