The diagram shows part of the curve $y = \frac{7}{6x + 1}$ intersecting the $y$-axis at $A(0, 7)$.
The tangent to the curve at the point $K(1, 1)$ intersects the $y$-axis at $B$.

1. Find the derivative of the curve equation

$y = 7(6x + 1)^{-1}$
$\frac{dy}{dx} = -7(6x + 1)^{-2} \cdot 6 = -\frac{42}{(6x + 1)^2}$

2. Calculate the gradient of the tangent at $K(1, 1)$

At $x = 1$,
$m = -\frac{42}{(6(1) + 1)^2} = -\frac{42}{49} = -\frac{6}{7}$

3. Find the equation of the tangent line and the coordinates of $B$

Equation of tangent at $K(1, 1)$:
$y - 1 = -\frac{6}{7}(x - 1)$
Since $B$ is the $y$-intercept, substitute $x = 0$:
$y - 1 = -\frac{6}{7}(0 - 1)$
$y = 1 + \frac{6}{7} = \frac{13}{7}$
Coordinates of $B$ are $\left(0, \frac{13}{7}\right)$.

1. Find the equation of the tangent and the coordinates of B from part (a)

$y = 7(6x+1)^{-1}$
$\frac{dy}{dx} = -7(6x+1)^{-2}(6) = -\frac{42}{(6x+1)^2}$
At $K(1, 1)$, $m = -\frac{42}{7^2} = -\frac{42}{49} = -\frac{6}{7}$
Equation of tangent: $y - 1 = -\frac{6}{7}(x - 1) \implies y = -\frac{6}{7}x + \frac{13}{7}$
At $x = 0$, $y = \frac{13}{7}$. So $B(0, \frac{13}{7})$.

2. Set up the integral for the area under the curve from x = 0 to x = 1

Area under curve $= \int_0^1 \frac{7}{6x+1} \, dx$

3. Evaluate the integral

$\int_0^1 \frac{7}{6x+1} \, dx = \left[ \frac{7}{6} \ln(6x+1) \right]_0^1$
$= \frac{7}{6} \ln(7) - \frac{7}{6} \ln(1)$
$= \frac{7}{6} \ln 7$

4. Calculate the area of the trapezium under the tangent KB

Area of trapezium $= \frac{1}{2} \times (OB + y_K) \times h$
$= \frac{1}{2} \times \left(\frac{13}{7} + 1\right) \times 1$
$= \frac{1}{2} \times \frac{20}{7} = \frac{10}{7}$

5. Find the shaded area by subtracting the trapezium area from the area under the curve

Shaded Area $= \frac{7}{6} \ln 7 - \frac{10}{7}$
$\approx 2.2702 - 1.4286 = 0.8416 \approx 0.842$ units$^2$