A-level Mathematics/OCR/C2/Integration/Solutions

Worked Solutions

1a)

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

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

We get:

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

b)

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

Again using our rule, we would get:

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

2a)

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

Using our rule, the intergral becomes

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

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

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

Therefore C = 3

b)

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

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

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

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

$2=8+14+0.2+C$

$C=-20.2$

3a)

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

Evaluating this we get:

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

Substituting in values we get:

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

$=4$

b)

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

Evaluating this we get:

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

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

$\approx 8.08$

4)

The question is simply to evaluate this definite integral:

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

5 )

$\int tan^{4}A$