Iterated Integrals — Question 6

Question 6

For a real parameter $a$, define $$I(a)=\int_0^2\int_{-1}^{1}\bigl(axy^2+x^2-2y\bigr)\,dy\,dx.$$

Tasks

1. Evaluate $I(a)$ exactly.

2. Find the unique value of $a$ for which the integrand’s average value on the rectangle is $1$.

3. Verify the resulting integral directly from the inner accumulation function.

Question 6 – Solution

Strategy. Use parity in the symmetric $y$-interval, retain the parameter symbolically, and convert the prescribed average into a total integral.

Step 1: Inner accumulation Since $-2y$ is odd, $$\begin{align*} \int_{-1}^{1}(axy^2+x^2-2y)\,dy &=ax\left[\frac{y^3}{3}\right]_{-1}^{1}+x^2[y]_{-1}^{1}\\ &=\frac{2ax}{3}+2x^2. \end{align*}$$ Thus $$I(a)=\int_0^2\left(\frac{2ax}{3}+2x^2\right)dx =\frac{4a}{3}+\frac{16}{3}.$$

Step 2: Prescribed average The rectangle’s area is $2\cdot 2=4$. Average value $1$ therefore requires $I(a)=4$: $$\frac{4a+16}{3}=4 \quad\Longrightarrow\quad \boxed{a=-1}.$$

Step 3: Direct verification With $a=-1$, the inner accumulation is $$-\frac{2x}{3}+2x^2.$$ Consequently, $$\int_0^2\left(-\frac{2x}{3}+2x^2\right)dx =\left[-\frac{x^2}{3}+\frac{2x^3}{3}\right]_0^2 =-\frac 43+\frac{16}{3}=4.$$ Dividing by area $4$ returns the required average $1$.