Edexcel A-Level Mathematics A2 P4.6.5 Use Integration to Find the Area QuestionsPractise finding areas from parametric equations by forming y dx/dt integrals and evaluating exact results without sketching.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsform area integrals from parametric equations using y dx/dt and correct t-limitsuse substitution or parts to evaluate exact parametric areas
Edexcel A-Level Mathematics A2 P4.6.5 Use Integration to Find the Area Questions question 1[Maximum number: 11]Figure 2Figure 2 shows a sketch of the curve defined by the parametric equationsx=t2+2ty=2t(3−t)a⩽t⩽bx=t^{2}+2 t \quad y=\frac{2}{t(3-t)} \quad a \leqslant t \leqslant bx=t2+2ty=t(3−t)2a⩽t⩽bwhere a and b are constants.The ends of the curve lie on the line with equation y=1Question (a)(a)Show that the area of region R is given byM−k∫abt+1t(3−t)dtM-k \int_{a}^{b} \frac{t+1}{t(3-t)} \mathrm{d} tM−k∫abt(3−t)t+1dtwhere M and k are constants to be found.[ 5 ]Show AnswerArea under curve =∫t=11t="2"y dx=∫12y dx dt dt=∫122t(3−t)×(2t+2)dt=\int_{t=11}^{t=" 2 "} y \mathrm{~d} x=\int_{1}^{2} y \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t=∫t=11t="2"y dx=∫12y dt dx dt=∫12t(3−t)2×(2t+2)dtt=1⇒x=3,t=2⇒x=8t=1 \Rightarrow x=3, \quad t=2 \Rightarrow x=8t=1⇒x=3,t=2⇒x=8So area of R=1×(8−3)−∫122t(3−t)×(2t+2)dtR=1 \times(8-3)-\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} tR=1×(8−3)−∫12t(3−t)2×(2t+2)dt=5−4∫12t+1t(3−t)dt=5-4 \int_{1}^{2} \frac{t+1}{t(3-t)} \mathrm{d} t=5−4∫12t(3−t)t+1dt(5)Question (b)(b)Use algebraic integration to find the exact area of R, giving your answer in simplest form.[ 6 ]Show Answer∫t+1t(3−t) dt=∫(13t+43(3−t)) dt=13lnt−43ln(3−t)\int\frac{t+1}{t(3-t)}\,dt =\int\left(\frac1{3t}+\frac4{3(3-t)}\right)\,dt =\frac13\ln t-\frac43\ln(3-t)∫t(3−t)t+1dt=∫(3t1+3(3−t)4)dt=31lnt−34ln(3−t)Area=5−4[13lnt−43ln(3−t)]12=5−4(13ln2+43ln2)\begin{aligned} \text{Area} &=5-4\left[\frac13\ln t-\frac43\ln(3-t)\right]_1^2\\ &=5-4\left(\frac13\ln2+\frac43\ln2\right) \end{aligned}Area=5−4[31lnt−34ln(3−t)]12=5−4(31ln2+34ln2)=5−203ln2=5-\frac{20}{3}\ln2=5−320ln2Add to Test
Question (a)(a)Show that the area of region R is given byM−k∫abt+1t(3−t)dtM-k \int_{a}^{b} \frac{t+1}{t(3-t)} \mathrm{d} tM−k∫abt(3−t)t+1dtwhere M and k are constants to be found.[ 5 ]Show AnswerArea under curve =∫t=11t="2"y dx=∫12y dx dt dt=∫122t(3−t)×(2t+2)dt=\int_{t=11}^{t=" 2 "} y \mathrm{~d} x=\int_{1}^{2} y \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t=∫t=11t="2"y dx=∫12y dt dx dt=∫12t(3−t)2×(2t+2)dtt=1⇒x=3,t=2⇒x=8t=1 \Rightarrow x=3, \quad t=2 \Rightarrow x=8t=1⇒x=3,t=2⇒x=8So area of R=1×(8−3)−∫122t(3−t)×(2t+2)dtR=1 \times(8-3)-\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} tR=1×(8−3)−∫12t(3−t)2×(2t+2)dt=5−4∫12t+1t(3−t)dt=5-4 \int_{1}^{2} \frac{t+1}{t(3-t)} \mathrm{d} t=5−4∫12t(3−t)t+1dt(5)
Question (b)(b)Use algebraic integration to find the exact area of R, giving your answer in simplest form.[ 6 ]Show Answer∫t+1t(3−t) dt=∫(13t+43(3−t)) dt=13lnt−43ln(3−t)\int\frac{t+1}{t(3-t)}\,dt =\int\left(\frac1{3t}+\frac4{3(3-t)}\right)\,dt =\frac13\ln t-\frac43\ln(3-t)∫t(3−t)t+1dt=∫(3t1+3(3−t)4)dt=31lnt−34ln(3−t)Area=5−4[13lnt−43ln(3−t)]12=5−4(13ln2+43ln2)\begin{aligned} \text{Area} &=5-4\left[\frac13\ln t-\frac43\ln(3-t)\right]_1^2\\ &=5-4\left(\frac13\ln2+\frac43\ln2\right) \end{aligned}Area=5−4[31lnt−34ln(3−t)]12=5−4(31ln2+34ln2)=5−203ln2=5-\frac{20}{3}\ln2=5−320ln2