OMPT integration basics practice: lesson and 10 problems

Answer: 4

An antiderivative is $x^3-x^2$. Evaluating between the bounds: $(8-4)-(0-0)=4$.

Answer: B — $\sin x$

I ask: whose derivative is $\cos x$? Differentiating $\sin x$ gives exactly $\cos x$, so $\sin x$ works. The minus signs live on the other pairing: $\int \sin xdx=-\cos x+C$.

Answer: 2

I rewrite the integrand as $x^{-1/2}$; its antiderivative is $\frac{x^{1/2}}{1/2}=2\sqrt{x}$. Then $2\sqrt{4}-2\sqrt{1}=4-2=2$.

Answer: C — $\frac{1}{3}{e}^{3x}+C$

Differentiating $e^{3x}$ produces an extra factor 3 by the chain rule, so integrating must compensate with $\frac{1}{3}$: the answer is $\frac{1}{3}{e}^{3x}+C$. Check by differentiating: $\frac{1}{3}\cdot 3e^{3x}=e^{3x}$.

Answer: Step 1

The antiderivative of $\sin x$ is $-\cos x$, not $\cos x$ — differentiating $\cos x$ gives $-\sin x$. With the sign fixed, the integral is $(-\cos \pi )-(-\cos 0)=1+1=2$. Steps 2 and 3 evaluate faithfully; the wrong seed in Step 1 flipped the final sign. An area under a curve that lies above the axis can never be negative, which is the giveaway.

Answer: 1.33

The curves meet where $x^2=2x$, at $x=0$ and $x=2$, and between them the line lies above the parabola (test $x=1$: $2>1$). So the area is ${\int }_0^2(2x-x^2)dx={\left[x^2-\frac{x^3}{3}\right]}_0^2=4-\frac{8}{3}=\frac{4}{3}\approx 1.33$.

Answer: 13

I rewrite the second term as $-2x^{-2}$ before doing anything else. Term by term: $6x^2\to 2x^3$ and $-2x^{-2}\to -2\cdot \frac{x^{-1}}{-1}=\frac{2}{x}$, so $F(x)=2x^3+\frac{2}{x}$. Differentiating back gives $6x^2-\frac{2}{x^2}$, so the antiderivative is right. Then $F(2)=16+1=17$ and $F(1)=2+2=4$, and the integral is $17-4=13$. Plausibility: the integrand runs from 4 at $x=1$ to 23.5 at $x=2$ over a width of 1, and 13 sits between those.

Answer: C — $6\sqrt{x}+\frac{1}{2}\cos (2x)+C$

Two separate jobs. First, $\frac{3}{\sqrt{x}}=3x^{-1/2}$, which integrates to $3\cdot \frac{x^{1/2}}{1/2}=6\sqrt{x}$. Second, $\int \sin (2x)dx=-\frac{\cos (2x)}{2}$, and the minus in front of the sine flips that to $+\frac{1}{2}\cos (2x)$. Together: $6\sqrt{x}+\frac{1}{2}\cos (2x)+C$. I check by differentiating: $6\cdot \frac{1}{2}{x}^{-1/2}=\frac{3}{\sqrt{x}}$ and $\frac{1}{2}\cdot (-\sin (2x))\cdot 2=-\sin (2x)$. Both pieces return.

Answer: 3

I treat $a$ as a number and integrate: ${\left[x^3+x^2\right]}_0^a=a^3+a^2$. So the condition is $a^3+a^2=36$. Trying small integers, $a=3$ gives $27+9=36$, so $a=3$ works. Is it the only positive solution? Dividing out the factor: $a^3+a^2-36=(a-3)(a^2+4a+12)$, and the quadratic has discriminant $16-48<0$, so it has no real roots. The answer is $a=3$. Check by integrating directly: ${\int }_0^3(3x^2+2x)dx=27+9=36$.

Answer: 23

Distance is the integral of speed, not of velocity, so I first find where the velocity changes sign: $12-3t^2=0$ at $t=2$. On $[0,2]$ the particle moves forward, on $[2,3]$ backward. Antiderivative: $F(t)=12t-t^3$. Forward leg: $F(2)-F(0)=24-8=16$. Backward leg: $F(3)-F(2)=(36-27)-16=-7$, so 7 metres in the other direction. Total distance: $16+7=23$. The displacement, which is what ${\int }_0^{3}v(t)dt$ gives, is $9$, and that is the number the question did not ask for.