Q23: Find $\int\frac{(x - 3)}{(x - 1)^3}e^x\,dx$⁠, or the shaded area beside $x^2 = y$

The integrand is $e^x$ times a fraction. That suggests the special form $e^x\big(f(x) + f'(x)\big)$⁠, so we try to split the fraction into a function plus its derivative.

Writing $x - 3 = (x - 1) - 2$ makes the denominator's factor appear on top. Then the fraction splits into two:

Cancel one factor $x - 1$ in the first part:

and $-\frac{2}{(x - 1)^3}$ is left over.

Write $f(x) = (x - 1)^{-2}$ and use the power rule:

(The $\cdot\,1$ is the derivative of the inside, $x - 1$⁠.)

This is exactly the second term, so the integrand really is $e^x\big(f(x) + f'(x)\big)$⁠.

This standard result comes from the product rule:

So $e^xf(x)$ is an antiderivative of $e^x\big(f(x) + f'(x)\big)$⁠.

Watch out: it works only when the second function is exactly the derivative of the first, sign included.

The region is bounded on the left by the y-axis $(x = 0)$ and on the right by the curve. It runs from $y = 0$ at O up to the line $y = 4$⁠.

So we use thin horizontal strips: each strip has length $x$ and width $dy$⁠, and we add them from $y = 0$ to $y = 4$⁠.

On the right arm of the parabola $x \ge 0$⁠, so $x^2 = y$ gives $x = \sqrt y$⁠, not $-\sqrt y$⁠.

Power rule, with $\sqrt y = y^{1/2}$⁠:

(Dividing by $\frac32$ is the same as multiplying by $\frac23$⁠.)

At $y = 4$⁠: $4^{3/2} = (\sqrt4)^3 = 2^3 = 8$⁠. At $y = 0$ the bracket is 0.

Then $\frac23 \times 8 = \frac{2 \times 8}{3} = \frac{16}{3}$⁠.

The region lies between the curve (above) and the x-axis (below), from the vertex at $x = 0$ to the line $x = 3$⁠. So we use thin vertical strips of height $y$ and width $dx$⁠.

In the first quadrant $y \ge 0$⁠, so $y^2 = x$ gives $y = \sqrt x$⁠.

As before, $\int x^{1/2}\,dx = \frac23x^{3/2}$⁠.

At $x = 3$⁠: $3^{3/2} = 3 \times \sqrt3 = 3\sqrt3$⁠. Then $\frac23 \times 3\sqrt3 = 2\sqrt3$⁠.