See answer given.

If we integrate each term separately, then \(t\) is relatively simple to integrate using the power rule; add one to the power and divide by the new power. This gives \(\frac{t^{2}}{2}\).

For \(-\frac{3}{t^{4}}\), we start by turning it into index form before integrating so that we can use the power rule.

\(\frac{3}{t^{4}}=3t^{-4}\)

Then perform the integration: \(\int_{ }^{ }3t^{-4}\ dt=\frac{3t^{-3}}{-3}+c=-t^{-3}+c\)

We finalise by turning the answer back into fractional form: \(-t^{-3}+c=-\frac{1}{t^{3}}+c\)

Putting this all together we obtain: \(\frac{t^{2}}{2}-\frac{1}{t^{3}}+c\)

Integrating the derivative gives the original function. See the answer given.

Before integrating, expand the bracket and simplify:

\(\left(x+4\right)^{2}-2=\left(x+4\right)\left(x+4\right)-2=x^{2}+8x+16-2=x^{2}+8x+14\)

Hence, \(\int_{ }^{ }x^{2}+8x+14\ dx=\frac{x^{3}}{3}+4x^{2}+14x+c\)

First integrate \(g'\left(x\right)\) to obtain \(g\left(x\right)=5x-\frac{4x^{3}}{3}+c\).

Then substitute the boundary condition values into \(g\left(x\right)\)to obtain the answer shown.

Start by integrating \(f'\left(x\right)\) to obtain \(f\left(x\right)=\frac{mx^{2}}{2}+4x+c\).

Notice that we have two boundary conditions (i.e. two separate coordinates) and two unknowns in the equation, \(m\) and \(c\). Substitute each coordinate into \(f\left(x\right)\) to find each of these unknowns.

Substituting \(\left(0,3\right)\) into \(f\left(x\right)\) gives \(f\left(x\right)=\frac{mx^{2}}{2\ }+4x+3\).

Then substituting \(\left(-2,7\right)\) gives \(m=1\).

Input this definite integral into the GDC to obtain \(18.9\) (\(3\) s.f.)

Step 1: \(h=\frac{\left(3-0\right)}{2}=1.5\)

Step 2:

Step 3:

Area = \(\frac{1}{2}\cdot1.5\left(1+2\left(2.2247...\right)+2.7320...\right)=6.14\ unit^{2}\ \left(3.s.f\right)\)

The coordinates in table form are:

Hence, utilising the trapezoidal rule:

\(\frac{1}{2}\cdot1\left(3+2\left(3.5+5+7.5\right)+11\right)=23\ units^{2}\)

See answer given.