Definite Integration flash cards

1. 1.State the King's property of definite integrals.

   King's Property: $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$

   Hint: Replace $x$ by $a+b-x$

2. 2.What is the property of $\int_0^{2a} f(x)\,dx$ in terms of $f(2a-x)$?

   $\int_0^{2a} f(x)\,dx=2\int_0^a f(x)\,dx$ if $f(2a-x)=f(x)$, and $\int_0^{2a} f(x)\,dx=0$ if $f(2a-x)=-f(x)$

   Hint: Check symmetry about $x=a$

3. 3.Evaluate $\int_{-a}^{a} f(x)\,dx$ for even and odd $f(x)$.

   $\int_{-a}^{a} f(x)\,dx=2\int_0^a f(x)\,dx$ if $f(x)$ is even (i.e., $f(-x)=f(x)$); $\int_{-a}^{a} f(x)\,dx=0$ if $f(x)$ is odd (i.e., $f(-x)=-f(x)$)

   Hint: Use $f(-x)=\pm f(x)$

4. 4.How is a definite integral defined as the limit of a sum?

   $\int_a^b f(x)\,dx=\lim_{n\to\infty}h\sum_{r=0}^{n-1} f(a+rh)$, where $h=\dfrac{b-a}{n}$

   Hint: Riemann sum with equal subintervals

5. 5.State the Fundamental Theorem of Calculus (Newton-Leibniz formula) used to evaluate $\int_a^b f(x)\,dx$.

   $\int_a^b f(x)\,dx=F(b)-F(a)$, where $F(x)$ is any antiderivative of $f(x)$, i.e., $F'(x)=f(x)$

   Hint: Antiderivative evaluated at limits

6. 6.Evaluate $\int_0^{\pi/2} \dfrac{\sin^3 x}{\sin^3 x+\cos^3 x}\,dx$.

   Let $I$ be the integral. By King's property ($x\to \frac{\pi}{2}-x$), $I=\int_0^{\pi/2}\dfrac{\cos^3x}{\cos^3x+\sin^3x}dx$. Adding the two forms of $I$ gives $2I=\int_0^{\pi/2}1\,dx=\dfrac{\pi}{2}$, so $I=\dfrac{\pi}{4}$.

   Hint: Replace $x\to \pi/2-x$ and add the two integrals.

7. 7.Evaluate $\int_0^\pi \dfrac{x\sin x}{1+\cos^2 x}\,dx$.

   Using King's property ($x\to\pi-x$), $I=\int_0^\pi\dfrac{(\pi-x)\sin x}{1+\cos^2x}dx$, so $2I=\pi\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx$. Put $t=\cos x$: this reduces to $\pi\int_{-1}^1\dfrac{dt}{1+t^2}=\pi\cdot\dfrac{\pi}{2}$. Hence $2I=\dfrac{\pi^2}{2}$, giving $I=\dfrac{\pi^2}{4}$.

   Hint: Apply $x\to\pi-x$, add, then substitute $t=\cos x$.

8. 8.Evaluate $\int_0^{\pi} \dfrac{x\,dx}{a^2\cos^2 x+b^2\sin^2 x}$ for $a,b>0$.

   Since the denominator $f(x)$ satisfies $f(\pi-x)=f(x)$, King's property gives $I=\int_0^\pi(\pi-x)f(x)dx=\pi\int_0^\pi f(x)dx-I$, so $2I=\pi\int_0^\pi f(x)dx$. By symmetry $\int_0^\pi f(x)dx=2\int_0^{\pi/2}f(x)dx=2\cdot\dfrac{\pi}{2ab}=\dfrac{\pi}{ab}$. Thus $2I=\dfrac{\pi^2}{ab}$, so $I=\dfrac{\pi^2}{2ab}$.

   Hint: Use $x\to\pi-x$, then the standard $\int_0^{\pi/2}\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\frac{\pi}{2ab}$.

9. 9.Prove the identity $\int_0^{2a} x\,f(x)\,dx = a\int_0^{2a} f(x)\,dx$ whenever $f(2a-x)=f(x)$.

   Let $I=\int_0^{2a}xf(x)dx$. Substituting $x\to 2a-x$ gives $I=\int_0^{2a}(2a-x)f(2a-x)dx=\int_0^{2a}(2a-x)f(x)dx=2a\int_0^{2a}f(x)dx-I$. Hence $2I=2a\int_0^{2a}f(x)dx$, i.e. $I=a\int_0^{2a}f(x)dx$.

   Hint: Substitute $x\to 2a-x$ and use the symmetry of $f$.

10. 10.Using the identity $\int_0^{2a}xf(x)dx=a\int_0^{2a}f(x)dx$ (valid when $f(2a-x)=f(x)$), evaluate $\int_0^{2\pi}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}\,dx$.

    Here $f(x)=\dfrac{\sin^8x}{\sin^8x+\cos^8x}$ satisfies $f(2\pi-x)=f(x)$, so with $2a=2\pi$ the integral equals $\pi\int_0^{2\pi}f(x)dx$. Since $f$ has period $\pi$ and $f(\pi-x)=f(x)$, $\int_0^{2\pi}f\,dx=4\int_0^{\pi/2}f\,dx$; King's property on $[0,\pi/2]$ gives $\int_0^{\pi/2}f\,dx=\pi/4$. So $\int_0^{2\pi}f\,dx=\pi$, and the answer is $\pi\cdot\pi=$ $\pi^2$.

    Hint: First reduce over $[0,2\pi]$ to $\pi/2$ of the plain integral of $f$, using periodicity.

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