Polynomial Integration II
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Find the area enclosed between \(y=6-x-x^2\) and the \(x\)-axis.
Find roots
\(x^2+x-6=0\Rightarrow(x+3)(x-2)=0\Rightarrow x=-3,2\)
Integrate
\(\displaystyle\int_{-3}^2(6-x-x^2)\,dx=[6x- frac{x^2}{2}- frac{x^3}{3}]_{-3}^2\)
Evaluate
\((12-2- frac{8}{3})-(-18- frac{9}{2}+9)= frac{22}{3}+ frac{27}{2}= frac{44+81}{6}\)
Answer
\(\dfrac{125}{6}\) square units
Question 2
Find the value of \(k>0\) such that \(\displaystyle\int_0^k(4-x^2)\,dx=0\).
Integrate
\([4x- frac{x^3}{3}]_0^k=4k- frac{k^3}{3}=0\Rightarrow k(4- frac{k^2}{3})=0\)
Solve for \(k>0\)
\(k^2=12\)
Question 3
A particle moves so that its velocity is \(v(t)=t^2-5t+4\) m/s. Find the displacement and the total distance travelled for \(0\leq t\leq4\).
Displacement
\(\displaystyle\int_0^4(t^2-5t+4)\,dt=[ frac{t^3}{3}- frac{5t^2}{2}+4t]_0^4= frac{64}{3}-40+16=- frac{8}{3}\)
Roots of \(v\)
\((t-1)(t-4)=0\Rightarrow t=1,4\)
Distance
\(\displaystyle\int_0^1 v\,dt-\int_1^4 v\,dt= frac{11}{6}-(- frac{9}{2})= frac{11}{6}+ frac{27}{6}= frac{38}{6}\)
Answer
Displacement: \(- frac{8}{3}\) m; Distance: \( frac{19}{3}\) m
Question 4
The gradient of a curve is \(\dfrac{dy}{dx}=3x^2-6x\). Given the curve passes through \((3,2)\), find the equation of the curve. Hence find the \(x\)-coordinates of any stationary points.
Integrate
\(y=x^3-3x^2+C\); using \((3,2)\): \(27-27+C=2\Rightarrow C=2\)
Stationary points
\(3x^2-6x=3x(x-2)=0\Rightarrow x=0,2\)
Answer
\(y=x^3-3x^2+2\); stationary at \(x=0\) and \(x=2\)
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