IB Maths AA HL Advanced Integration Questions

1. Identify the inner function and set up the substitution: Let $u=x^2$.

2. Differentiate to find the differential relationship: $\frac{du}{dx}=2x⟹dx=\frac{du}{2x}$.

3. Calculate the new limits of integration in terms of $u$: When $x=0$, $u=0^2=0$. When $x=1$, $u=1^2=1$.

4. Substitute $u$, $dx$, and the new limits into the integral: ${\int }_0^{1}x{e}^u\frac{du}{2x}$.

5. Cancel the $x$ terms and factor out the constant: $\frac{1}{2}{\int }_0^1{e}^{u}du$.

6. Evaluate the definite integral using the anti-derivative: $\frac{1}{2}[e^u{]}_0^1=\frac{1}{2}(e^1-e^0)=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{e}\mathbf{-}\mathbf{1}\mathbf{)}$.

1. Select $u$ and $dv$ using the LATE hierarchy (Logarithmic before Algebraic): Let $u=\ln x$ and $dv=xdx$.

2. Differentiate $u$ and integrate $dv$: $du=\frac{1}{x}dx$ and $v=\frac{x^2}{2}$.

3. State the integration by parts formula: $\int udv=uv-\int vdu$.

4. Substitute the expressions into the formula: ${[\frac{x^2}{2}\ln x]}_1^e-{\int }_1^e\frac{x^2}{2}\cdot \frac{1}{x}dx$.

5. Simplify and evaluate the remaining integral: ${[\frac{x^2}{2}\ln x]}_1^e-{\int }_1^e\frac{x}{2}dx={[\frac{x^2}{2}\ln x-\frac{x^2}{4}]}_1^e$.

6. Substitute the limits to find the exact value: $(\frac{e^2}{2}(1)-\frac{e^2}{4})-(0-\frac{1}{4})=\frac{e^2}{4}+\frac{1}{4}=\frac{{\mathbf{e}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}{\mathbf{4}}$.

1. State the formula for a volume of revolution about the $x$-axis: $V=\pi {\int }_a^b{y}^{2}dx$.

2. Square the function and set up the integral: $V=\pi {\int }_3^5\frac{1}{x^2-4}dx$.

3. Factorise the denominator and split into partial fractions: $\frac{1}{(x-2)(x+2)}\equiv \frac{A}{x-2}+\frac{B}{x+2}$. Solving gives $A=\frac{1}{4}$ and $B=-\frac{1}{4}$.

4. Rewrite the integral using the partial fractions: $V=\frac{\pi }{4}{\int }_3^5(\frac{1}{x-2}-\frac{1}{x+2})dx$.

5. Integrate to log forms: $\frac{\pi }{4}[\ln |x-2|-\ln |x+2|{]}_3^5$.

6. Evaluate the limits and simplify using log laws: $\frac{\pi }{4}((\ln 3-\ln 7)-(\ln 1-\ln 5))=\frac{\pi }{4}(\ln \frac{3}{7}+\ln 5)=\frac{\pi }{\mathbf{4}}\ln \left(\frac{\mathbf{15}}{\mathbf{7}}\right)$.