Worked Solutions

1a)

${\displaystyle \int 2x^{5}\,dx}$

Using our rule: That ${\displaystyle \int {\frac {dy}{dx}}=x^{n}dx}$ is equal to ${\displaystyle y={\frac {x^{(n+1)}}{(n+1)}}+C}$

We get:

${\displaystyle y={\frac {2x^{6}}{6}}+C}$

b)

${\displaystyle \int 7x^{6}+2x^{3}-x^{2}\,dx}$

Again using our rule, we would get:

${\displaystyle y=x^{7}+{\frac {x^{4}}{2}}-{\frac {x^{3}}{3}}+C}$

2a)

${\displaystyle \int x+5\,dx}$ given that the point ${\displaystyle (0,3)}$ lies on the curve.

Using our rule, the intergral becomes

${\displaystyle y={\frac {x^{2}}{2}}+5x+C}$

Now we can sub in our points ${\displaystyle (0,3)}$, So that:

${\displaystyle 3={\frac {0^{2}}{2}}+5(0)+C}$

Therefore C = 3

b)

${\displaystyle \int 3x^{2}+7x+0.1\,dx}$

Evaulating this we get: ${\displaystyle x^{3}+{\frac {7x^{2}}{2}}+0.1x+C}$

Given (2,2), subing these points in:

${\displaystyle 2=2^{3}+{\frac {7(2^{2})}{2}}+0.2+C}$

${\displaystyle 2=8+14+0.2+C}$

${\displaystyle C=-20.2}$

3a)

${\displaystyle \int _{0}^{2}x+1\,dx}$

Evaluating this we get:

${\displaystyle {\Bigg \lfloor }{\frac {x^{2}}{2}}+x{\Bigg \rceil }_{0}^{2}}$

Substituting in values we get:

${\displaystyle {\Bigg \lfloor }\left({\frac {2^{2}}{2}}+2\right)-\left({\frac {0^{2}}{2}}+0\right){\Bigg \rceil }}$

${\displaystyle =4}$

b)

${\displaystyle \int _{-3}^{4.7}{\frac {1}{7}}x^{\frac {1}{3}}+1\,dx}$

Evaluating this we get:

${\displaystyle {\Bigg \lfloor }{\frac {3x^{\frac {4}{3}}}{28}}+x{\Bigg \rceil }_{-3}^{4.7}}$

${\displaystyle {\Bigg \lfloor }\left({\frac {3(4.7)^{\frac {4}{3}}}{28}}+4.7\right)-\left({\frac {3(-3)^{\frac {4}{3}}}{28}}-3\right){\Bigg \rceil }}$

${\displaystyle \approx 8.08}$

4)

The question is simply to evaluate this definite integral:

${\displaystyle {\begin{aligned}\int _{-2}^{0}{\bigg (}y=-x^{4}-{\frac {1}{2}}x^{3}+3x^{2}{\bigg )}\,dx&={\Bigg \lfloor }\left({\frac {-1}{5}}x^{5}-{\frac {1}{8}}x^{4}+x^{3}\right){\Bigg \rceil }_{-2}^{0}\\&=-{\bigg (}{\frac {-1}{5}}*-2^{5}-{\frac {1}{8}}*-2^{4}-2^{3}{\bigg )}\\&=3.6\end{aligned}}}$

5 )

${\displaystyle \int tan^{4}A}$