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 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)
 * $$ \int 3x^2 + 7x +0.1 \,dx $$


 * 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{align} \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{align} $$

Integration/Solutions