Detailed Answer

a) \(\int{4x + 1} \ dx = \frac{4x^2}{2} + x = 2x^2 + x + c\)

b) \(\int{5x^3 + 6x^2 + 10} \ dx = \frac{5x^4}{4} + \frac{6x^3}{3} + 10x = \frac{5}{4}x^4 + 2x^3 + 10x + c\)

c) \(\int{\frac{1}{2}x^2 + 2x} \ dx = \frac{1}{2} * \frac{x^3}{3} + \frac{2x^2}{2} = \frac{1}{6}x^3 + x^2 + c\)

d) \(\int{\frac{2x + 3x^2}{x}} \ dx = \int{2 + 3x} \ dx = 2x + \frac{3x^2}{2} + c\)

e) \(\int{3x(2x+1)(4-3x)} \ dx = \int{3x(8x - 6x^2 + 4 - 3x)} \ dx = \int{15x^2 - 18x^3 + 12x} \ dx = \frac{15x^3}{3} - \frac{18x^4}{4} + \frac{12x^2}{2} = 5x^3 - \frac{9}{2}x^4 + 6x^2 + c\)

Detailed Answer

[Maximum mark: 10]

a) To find the equation of the tangent to \( f(x) \) at point \( Q \), we first need to find the gradient at \( Q \).

\[f'(x) = 9x^2 - 4x + 1\]

Substituting \( x = 2 \) into the derivative to find the gradient at \( Q \):

\[f'(2) = 9(2)^2 - 4(2) + 1 = 36 - 8 + 1 = 29\]

So, the slope of the tangent line at \( Q \) is 29.

The equation of the tangent line in point-slope form is:

\[y - y_1 = m(x - x_1)\]

Here, \( m = 29 \) and the point \((x_1, y_1) = (2, 14)\). Substituting these values:

\[y - 14 = 29(x - 2)\]

\[y - 14 = 29x - 58\]

\[y = 29x - 44\]

b) To find the equation of \( f(x) \), we need to integrate \( f'(x) = 9x^2 - 4x + 1 \).

\[f(x) = \int (9x^2 - 4x + 1) \, dx\]

\[f(x) = 3x^3 - 2x^2 + x + C\]

We need to find the constant \( C \) using the point \( Q(2, 14) \):

\[f(2) = 14\]

Substituting \( x = 2 \) into the equation:

\[14 = 3(2)^3 - 2(2)^2 + 2 + C\]

\[14 = 3(8) - 2(4) + 2 + C\]

\[14 = 24 - 8 + 2 + C\]

\[14 = 18 + C\]

\[C = 14 - 18\]

\[C = -4\]

Thus, the equation of \( f(x) \) is:

\[f(x) = 3x^3 - 2x^2 + x - 4\]

Detailed Answer

a) This will simply be an integral over the time interval \([0,t]\):

\[\int_{0}^{t}{100 - 0.3t \, dt}\]

b) Here we can use what we got from part (a) and plug in the value of 20:

\[\int_{0}^{20} 100 - 0.3t \, dt = \left[ 100t - \frac{0.3t^2}{2} \right]_0^{20} = (100 \cdot 20 - \frac{0.3 \cdot 20^2}{2}) - 0 = 1940\]

So, in the first 20 seconds \(1940 \, \text{cm}^3\) were filled.

c) Now, we have to equate the integral to the value of 10000:

\[\int_{0}^{t} 100 - 0.3t \, dt = 10000\]

\[\left[ 100t - \frac{0.3t^2}{2} \right]_0^{t} = 10000\]

\[100t - \frac{0.3t^2}{2} - 0 = 10000\]

\[100t - \frac{0.3t^2}{2} = 10000\]

This can be solved easily on a GDC by plotting two curves:

Naturally, the first point will be the one that interests us because that's when \(10000 \, \text{l}\) will be first obtained, so for the value of 122.515 seconds. By rounding it to the nearest minute we get the answer of 2 minutes.

Detailed Answer

a) We first need to integrate the function for the cost change:

\[\int{25 - \frac{b}{10}} \, db = 25b - \frac{b^2}{20} + C\]

Now, we can obtain the constant \(C\) by plugging the given point \((1500, 10000)\).

\[25 \cdot 1500 - \frac{1500^2}{20} + C = 10000\]

\[C = 85000\]

So, the final answer is:

\[C = 25b - \frac{b^2}{20} + 85000\]

b) The fixed costs can be found when there is no bread being baked, so when \(b = 0\). By plugging it into the costs function we can clearly see that the fixed costs are equal to 85000.

c) This can be easiest seen by plotting the cost function on the GDC. By doing that we can clearly see that the maximum is when \(b = 250\) (rounded to 3 significant figures).

Detailed Answer

a) This part can be easily solved by plotting the graph on the GDC:

So, the local maximum occurs when \(x = 1.36\)

b) This question again can be easily done in GDC by calculating two \(x\)-intercepts for the segment of our interest. The first intercept is for when \(x = 0\), and the second one when \(x = 2.5\).

c) Now, to find the area under the curve we have to calculate the integral. However, this can be done faster and easier in a GDC (especially for a complex function like this one):

So, the area of the shaded region is 37.76 (rounded to two decimal points).

Detailed Answer

a) By selecting an \(x\)-window on the GDC from 0 to 2.5, and also choosing relatively low \(y\)-maximum values, we get:

b) To find the derivative of this function we first have to get rid of the bracket

\[f(x) = -\frac{x}{3}(x-2) = -\frac{x^2}{3} + \frac{2x}{3}\]

\[\frac{dy}{dx}= -\frac{2x}{3} + \frac{2}{3}\]

c) This will simply be the area enclosed by the values 0 and 2, so:

\[\int_{0}^{2} -\frac{x}{3}(x-2) \, dx\]

d) To calculate it we can either get the integral manually, or input it into the GDC (which is a faster and easier option):

So, the area \(Z\) is equal to 0.444.

Detailed Answer

a) To find the equation of \( f(x) \), we integrate the given gradient function \( f'(x) = kx + 5 \):

\[f'(x) = kx + 5\]

Integrating \( f'(x) \) with respect to \( x \):

\[f(x) = \int (kx + 5) \, dx = \frac{kx^2}{2} + 5x + C\]

Given that \( f(x) \) cuts the y-axis at \( y = 2 \), we know \( f(0) = 2 \):

\[f(0) = \frac{k \cdot 0^2}{2} + 5 \cdot 0 + C = 2 \implies C = 2\]

So, the equation becomes:

\[f(x) = \frac{kx^2}{2} + 5x + 2\]

Given that \( f(x) \) cuts the x-axis at \( x = 2 \), we know \( f(2) = 0 \):

\[f(2) = \frac{k \cdot 2^2}{2} + 5 \cdot 2 + 2 = 0 \implies 2k + 10 + 2 = 0 \implies 2k + 12 = 0 \implies k = -6\]

Thus, the function \( f(x) \) is:

\[f(x) = \frac{-6x^2}{2} + 5x + 2 = -3x^2 + 5x + 2\]

b) This can be easily found in the GDC:

\[\text{Area} = \int_{0}^{2} (-3x^2 + 5x + 2) \, dx = 6\]

c) To find the tangent at \( x = 1 \), we need the slope at \( x = 1 \). The slope is given by \( f'(x) \):

\[f'(x) = kx + 5 = -6x + 5\]

Evaluating the slope at \( x = 1 \):

\[f'(1) = -6 \cdot 1 + 5 = -1\]

The slope of the tangent line at \( x = 1 \) is \( -1 \).

Next, we find the y-coordinate at \( x = 1 \):

\[f(1) = -3(1)^2 + 5 \cdot 1 + 2 = -3 + 5 + 2 = 4\]

So the point of tangency is \( (1, 4) \).

Using the point-slope form of the equation of a line:

\[y - y_1 = m(x - x_1)\]

Where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency:

\[y - 4 = -1(x - 1)\]

Simplifying:

\[y - 4 = -x + 1 \implies y = -x + 5\]

Detailed Answer

a) To solve this part we have to apply the trapezoidal rule formula:

\[A = \frac{1}{2}h((y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1}))\]

When it comes to \(y\)-values, we can easily get them from the table. To calculate h we also have to apply the formula:

\[h = \frac{b-a}{n}\]

Where \(b\) and \(a\) are the upper and lower \(x\)-boundaries, so:

\[h = \frac{4 - 2}{4} = \frac{1}{2}\]

Now, by plugging all values into the first formula:

\[A = \frac{1}{4} \times ((0 + 52) + 2(8.5 + 20 + 34.5)) = 44.5\]

b) Now, to find the exact area we simply input the formula and the boundaries into the GDC:

The total exact area is equal to 44.

c) Now, using the formula for percentage error, we get that:

\[ \varepsilon = \left|\frac{v_a - v_e}{v_e}\right| \cdot 100\% = \left|\frac{44.5 - 44}{44}\right| \cdot 100\% = 1.14%\%\]

d) To find the derivative, we should first multiply out the equation we have for the function:

\[2(3x+1)(x-2) = 2(3x^2 - 6x + x - 2) = 6x^2 - 10x - 4\]

\[f'(x) = 12x - 10\]

Detailed Answer

a) To find the equation of the tangent to \( f(x) = -2x^3 + 6x^2 \) at \( x = 3 \) we have to first find the derivative:

\[ f(x) = -2x^3 + 6x^2 \]

\[ f'(x) = \frac{d}{dx}(-2x^3 + 6x^2) = -6x^2 + 12x \]

\[ f'(3) = -6(3)^2 + 12(3) = -54 + 36 = -18 \]

So, the slope of the tangent at \( x = 3 \) is \(-18\).

Now, we find the y-coordinate at \( x = 3 \):

\[ f(3) = -2(3)^3 + 6(3)^2 = -54 + 54 = 0 \]

So, the point of tangency is \( (3, 0) \).

Then, we find the tangent line:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = -18 \), \( x_1 = 3 \), and \( y_1 = 0 \):

\[ y - 0 = -18(x - 3) \implies y = -18x + 54 \]

b) To find the shaded area we first have to find the x-intercepts:

\[ f(x) = -2x^3 + 6x^2 = x^2(-2x + 6) \]

\[ x^2(-2x + 6) = 0\]

\[ x = 0, x = 3\]

Now, we can plug the function into the GDC and find the area:

So, the area is equal to 13.5.

c)

\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} \]

Given the vertices \( A(1, 0) \), \( B(5, 0) \), and \( C(3, y) \), the base \( AB \) is \( 5 - 1 = 4 \).

\[ \frac{1}{2} \times 4 \times y = 13.5 \]

\[ 2y = 13.5 \implies y = \frac{13.5}{2} = 6.75 \]

The y-coordinate of point \( C \) is \( 6.75 \).

Detailed Answer

a) To find the derivative we need to first multiply out the initial formula:

\[f(x) = 2x(x-10)\left(\frac{1}{2}x - 5\right) = 2x\left(\frac{1}{2}x^2 - 5x - 5x + 50\right) = 2x\left(\frac{1}{2}x^2 - 10x + 50\right)\]

\[f(x) = x^3 - 20x^2 + 100x\]

\[f'(x) = 3x^2 - 40x + 100\]

b) Maximum height can either be found by equalizing the derivative to 0, or by plugging the function into the GDC.

As it can be seen, the maximum height (y-value) is 148.

c) Again, this can be quickly solved in the GDC by finding the zero of this function, and as it can be seen from the figure below, it happens when \(x = 10\). So \(B\) has coordinates \((10,0)\).

d) So, we are looking to use the trapezoidal rule in the interval \(0 < x < 10\), with 5 strips. Therefore, we know that each strip will be 2 units wide (10 divided by 5). By looking at the formula:

\[A = \frac{1}{2}h((y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1}))\]

We can see that we are missing the corresponding \(y\)-values to each strip. Let's then create a table and fill them in:

So, our function has to be filled in with all of our \(x\)-values, for example:

\[f(0) = 2 \cdot 0(0-10)\left(\frac{1}{2}0 - 5\right) = 0\]

\[f(2) = 2 \cdot 2(2-10)\left(\frac{1}{2}2 - 5\right) = 128\]

All other values are filled in respectively, so we get:

Now, the last value we still need is \(h\), which can be calculated using the formula:

\[h = \frac{b-a}{n}= \frac{10-0}{5} = 2\]

So, by inputting all values into the formula we get:

\[A = \frac{1}{2} \cdot 2((0 + 0) + 2(128 + 144 + 96 + 32)) = 800\]

e)

\[\int_{0}^{10} 2x(x-10)\left(\frac{1}{2}x - 5\right) \, dx\]

f) By plugging it into the GDC we get:

So, the exact area is equal to 833.

g) Using the formula for percentage error, we get that:

\[\varepsilon = \left|\frac{v_a - v_e}{v_e}\right| \cdot 100\% = \left|\frac{800 - 833.33\ldots}{833.33\ldots}\right| \cdot 100\% = 4\%\]