Math Subject exam 9768 Q.4

Question

The question is given in the following picture:

Could anyone give me a hint please?

Answer 1

We can directly evaluate each integral and solve for the unknown constant $b$.

The following is given:

$$\int_0^bxdx=\int_0^bx^2dx$$

So:

$$\int_0^bxdx=\bigg[\frac {x^2}{2}\bigg]^b_0=\frac {b^2}{2}$$ $$\int_0^bx^2dx=\bigg[\frac {x^3}{3}\bigg ]^b_0=\frac {b^3}{3}$$

So equate each evaluation and solve for $b$:

$$\frac {b^3}{3}=\frac {b^2}{2}\implies 2b^3-3b^2=0\implies b^2(2b-3)=0$$ $$\therefore b=0, \frac 32$$

$b= 0$ is the trivial case, and thus the only relevant value is $b=\frac 32$.

The equivalent function can be determined by equating each separate equality and solving for its roots.

Thus we have:

$$x^2=x\implies x^2-x=0\implies x(x-1)=0$$

$$\therefore x=0,1$$

The area bounded by the curves $y=x$ and $y=x^2$, as denoted by the above graph, can be determined by taking the integral of $f(x)=x^2-x$ on the closed interval $[1,\frac 32]$.

$$\int^b_1(x^2-x)dx=\bigg[\frac {x^3}{3}-\frac {x^2}{2}\bigg]^{\frac 32}_1=\frac {(\frac 32)^3}{3}-\frac {(\frac 32)^2}{2}-\bigg(\frac 13-\frac 12\bigg)=\frac 98-\frac 98-\frac 13+\frac 12=\frac 16$$

As we can clearly see, this is equivalent to $\int_0^1(x-x^2)dx$. This is due to the initial condition stated at the very top.

Answer 2

Hint:

The shaded region is equal to

$$\int_0^1x-x^2dx$$

Can you identify the region that corresponds to that integral?

Answer 3

Hint: with these conditions, you can do it without finding the vertical line, and show the answer is equal to $$\int_0^1 x-x^2 dx$$