Integration, Worked Examples (KBAT)

First find where the curve and line intersect by setting them equal:

$$x^{2}=x+2 \;\Rightarrow\; x^{2}-x-2=0 \;\Rightarrow\; (x-2)(x+1)=0 \;\Rightarrow\; x=-1 \text{ or } x=2$$

Between these values the line lies above the curve, test $x=0$: the line gives $2$, the curve gives $0$. So the enclosed area is the integral of (upper − lower):

$$\text{Area}=\int_{-1}^{2}\big[(x+2)-x^{2}\big]\,dx=\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}$$

Evaluate at the upper limit $x=2$:

$$\frac{(2)^{2}}{2}+2(2)-\frac{(2)^{3}}{3}=2+4-\frac{8}{3}=\frac{10}{3}$$

Then at the lower limit $x=-1$:

$$\frac{(-1)^{2}}{2}+2(-1)-\frac{(-1)^{3}}{3}=\frac{1}{2}-2+\frac{1}{3}=-\frac{7}{6}$$

Subtract the lower value from the upper value:

$$\text{Area}=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$$

The volume generated by rotating a region about the x-axis is $V=\pi\displaystyle\int_{a}^{b}y^{2}\,dx$. Squaring the curve removes the root neatly:

$$y=\sqrt{x+1} \;\Rightarrow\; y^{2}=x+1$$

Substitute into the volume formula with limits $x=0$ and $x=3$:

$$V=\pi\int_{0}^{3}(x+1)\,dx=\pi\left[\frac{x^{2}}{2}+x\right]_{0}^{3}$$

Apply the limits; the lower limit contributes nothing:

$$=\pi\left(\frac{(3)^{2}}{2}+3\right)-\pi(0)=\pi\left(\frac{9}{2}+3\right)=\frac{15\pi}{2}$$

Part (a). Differentiate $\dfrac{x}{x+1}$ with the quotient rule $\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$, with $u=x$ and $v=x+1$, so that $\frac{du}{dx}=1$ and $\frac{dv}{dx}=1$:

$$\frac{d}{dx}\left(\frac{x}{x+1}\right)=\frac{(x+1)(1)-x(1)}{(x+1)^{2}}=\frac{x+1-x}{(x+1)^{2}}=\frac{1}{(x+1)^{2}}$$

This proves the required result. Part (b).

Because integration reverses differentiation, part (a) tells us that $\dfrac{x}{x+1}$ is an antiderivative of $\dfrac{1}{(x+1)^{2}}$. So apply the limits to it directly:

$$\int_{0}^{1}\frac{1}{(x+1)^{2}}\,dx=\left[\frac{x}{x+1}\right]_{0}^{1}=\frac{1}{1+1}-\frac{0}{0+1}=\frac{1}{2}-0=\frac{1}{2}$$

This region is bounded by the y-axis, not the x-axis, so integrate with respect to $y$ instead, first rewrite the equation of the curve with $x$ as the subject.

$$y=x^{2} \;\Rightarrow\; x=\sqrt{y} \quad (x\ge 0)$$

The area between a curve and the y-axis is $\displaystyle\int_{c}^{d}x\,dy$, so integrate $\sqrt{y}$ between $y=1$ and $y=4$:

$$\text{Area}=\int_{1}^{4}\sqrt{y}\,dy=\left[\frac{2}{3}y^{3/2}\right]_{1}^{4}$$

Evaluate at the limits:

$$=\frac{2}{3}(4)^{3/2}-\frac{2}{3}(1)^{3/2}=\frac{2}{3}(8)-\frac{2}{3}(1)=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$$

Displacement is a single definite integral of $v$, but total distance needs the particle's direction to be checked first, find when $v=0$.

$$v=0: \; 6-2t=0 \;\Rightarrow\; t=3$$

(a) The displacement from $O$ at $t=4$ is the definite integral of $v$ from $t=0$ to $t=4$:

$$s=\int_{0}^{4}(6-2t)\,dt=\big[6t-t^{2}\big]_{0}^{4}=(24-16)-0=8$$

(b) The particle reverses direction at $t=3$, so find the distance covered before and after that instant separately:

$$\int_{0}^{3}(6-2t)\,dt=\big[6t-t^{2}\big]_{0}^{3}=18-9=9 \qquad \int_{3}^{4}(6-2t)\,dt=\big[6t-t^{2}\big]_{3}^{4}=8-9=-1$$

The second integral is negative because the particle is moving backwards over that second, take its magnitude, then add both stages of the journey:

$$\text{Total distance}=9+|-1|=9+1=10$$

Rotating about the y-axis uses $V=\pi\displaystyle\int_{c}^{d}x^{2}\,dy$, so the curve needs to be written with $x^{2}$ as the subject.

$$y=x^{2} \;\Rightarrow\; x^{2}=y$$

Substitute directly into the volume formula between $y=0$ and $y=4$, no extra algebra is needed, since $x^{2}$ is already just $y$:

$$V=\pi\int_{0}^{4}y\,dy=\pi\left[\frac{y^{2}}{2}\right]_{0}^{4}$$

Apply the limits:

$$=\pi\left(\frac{16}{2}\right)-\pi(0)=8\pi$$

Integrate the second derivative once to get the gradient function, which introduces a constant of integration:

$$\frac{d^{2}y}{dx^{2}}=6x-4 \;\Rightarrow\; \frac{dy}{dx}=3x^{2}-4x+c_{1}$$

A turning point means the gradient is zero there, so substitute $x=1,\ \dfrac{dy}{dx}=0$ to find $c_{1}$:

$$3(1)^{2}-4(1)+c_{1}=0 \;\Rightarrow\; -1+c_{1}=0 \;\Rightarrow\; c_{1}=1$$

So $\dfrac{dy}{dx}=3x^{2}-4x+1$. Integrate a second time to get $y$, introducing a second constant:

$$y=x^{3}-2x^{2}+x+c_{2}$$

Substitute the point $(2,7)$ to find $c_{2}$:

$$7=(2)^{3}-2(2)^{2}+(2)+c_{2}=8-8+2+c_{2}=2+c_{2} \;\Rightarrow\; c_{2}=5$$

The curve dips below the x-axis between its two roots, so a single integral from $x=0$ to $x=4$ would let the negative part cancel some of the positive part. Split the region at $x=1$ and $x=3$ instead, and start from the antiderivative:

$$\int(x^{2}-4x+3)\,dx=\frac{x^{3}}{3}-2x^{2}+3x=F(x)$$

Evaluate $F$ at each boundary:

$$F(0)=0,\quad F(1)=\frac{1}{3}-2+3=\frac{4}{3},\quad F(3)=9-18+9=0,\quad F(4)=\frac{64}{3}-32+12=\frac{4}{3}$$

The curve is above the axis on $[0,1]$ and $[3,4]$, but below it on $[1,3]$, so take the magnitude of that middle piece before adding:

$$\text{Area}=\big[F(1)-F(0)\big]+\big|F(3)-F(1)\big|+\big[F(4)-F(3)\big]=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}$$

Add the three equal pieces:

$$\text{Area}=4$$