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)

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

Using our rule, the integral 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)

$\int 3x^2+7x+0.1dx$

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)

$\underset{0}{\overset{2}{\int }}x+1dx$

Evaluating this we get:

$\lfloor \frac{x^2}{2}+x{\rceil }_0^2$

Substituting in values we get:

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

$=4$

b)

$\underset{-3}{\overset{4.7}{\int }}\frac{1}{7}{x}^{\frac{1}{3}}+1dx$

Evaluating this we get:

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

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

$\approx 8.08$

4)

The question is simply to evaluate this definite integral:

$\begin{array}{cc}\underset{-2}{\overset{0}{\int }}(y=-x^4-\frac{1}{2}{x}^3+3x^2)dx& =\lfloor (\frac{-1}{5}{x}^5-\frac{1}{8}{x}^4+x^3){\rceil }_{-2}^0\\ & =-(\frac{-1}{5}*-2^5-\frac{1}{8}*-2^4-2^3)\\ & =3.6\end{array}$

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