IB Maths AI HL Advanced Integration and Kinematics Questions

1. Set up the definite integral for the area under the curve between the given roots: $\text{Area}={\int }_1^5(-x^2+6x-5)dx$.

2. Find the anti-derivative: ${[-\frac{x^3}{3}+\frac{6x^2}{2}-5x]}_1^5={[-\frac{x^3}{3}+3x^2-5x]}_1^5$.

3. Substitute the upper limit ($5$): $-\frac{125}{3}+3(25)-25=-\frac{125}{3}+50=\frac{25}{3}$.

4. Substitute the lower limit ($1$): $-\frac{1}{3}+3(1)-5=-\frac{1}{3}-2=-\frac{7}{3}$.

5. Subtract the evaluations to find the exact area: $\frac{25}{3}-(-\frac{7}{3})=\frac{32}{3}$. Area is $\frac{32}{3}$ square units.

1. Identify when the particle is momentarily at rest by setting $v(t)=0$: $t^2-7t+10=0⟹(t-2)(t-5)=0$. The particle rests at $t=2$ and $t=5$.

2. Recognize that the velocity is negative between $t=2$ and $t=5$, meaning the particle moves backwards. Total distance requires the absolute value of velocity.

3. Set up the total distance integral, splitting at the sign changes: $D={\int }_0^{2}vdt-{\int }_2^{5}vdt+{\int }_5^{6}vdt$.

4. Evaluate using the antiderivative $F(t)=\frac{t^3}{3}-\frac{7t^2}{2}+10t$: the three segments give $\frac{26}{3}$, $\frac{9}{2}$, and $\frac{11}{6}$ respectively.

5. Sum the three positive contributions: $\frac{26}{3}+\frac{9}{2}+\frac{11}{6}=\frac{52+27+11}{6}=\frac{90}{6}=15$.

6. The total distance travelled is exactly $15\text{ m}$.

1. Find the points of intersection to establish the limits: $\sqrt{x}=x^2⟹x=x^4⟹x(x^3-1)=0$. The limits are $x=0$ and $x=1$.

2. Identify the upper and lower curves on the interval $[0,1]$: $y=\sqrt{x}$ is above $y=x^2$.

3. Set up the volume of revolution integral using $V=\pi \int (y_{upper}^2-y_{lower}^2)dx$: $V=\pi {\int }_0^1((\sqrt{x}{)}^2-(x^2{)}^2)dx$.

4. Simplify the integrand algebraically: $V=\pi {\int }_0^1(x-x^4)dx$.

5. Integrate the polynomial: $V=\pi {[\frac{x^2}{2}-\frac{x^5}{5}]}_0^1$.

6. Evaluate the exact volume: $\pi (\frac{1}{2}-\frac{1}{5})=\pi (\frac{5}{10}-\frac{2}{10})=$ $\frac{3\pi }{10}$.