OMPT integration techniques practice: lesson and 10 problems

Answer: 3.75

The factor $2x$ is exactly the derivative of $x^2+1$, so I substitute $u=x^2+1$, $du=2xdx$. The bounds become $u=1$ to $u=2$, and the integral is ${\int }_1^2{u}^{3}du={\left[\frac{u^4}{4}\right]}_1^2=\frac{16-1}{4}=\frac{15}{4}=3.75$.

Answer: B — $(x-1)e^x+C$

Integration by parts with $u=x$ and $dv=e^{x}dx$ gives $x{e}^x-\int e^{x}dx=x{e}^x-e^x=(x-1)e^x+C$. Differentiating the answer confirms it: $e^x+(x-1)e^x=x{e}^x$.

Answer: 0.57

By parts with $u=x$, $dv=\cos xdx$: the antiderivative is $x\sin x+\cos x$. Evaluating: at $\frac{\pi }{2}$ it gives $\frac{\pi }{2}\cdot 1+0=\frac{\pi }{2}$; at 0 it gives $0+1=1$. The integral is $\frac{\pi }{2}-1\approx 0.57$.

Answer: C — $u=1-x^2$

With $u=1-x^2$, $du=-2xdx$, the stray $x$ in the numerator is absorbed: the integral becomes $-\frac{1}{2}\int u^{-1/2}du=-\sqrt{u}+C=-\sqrt{1-x^2}+C$. The presence of the inner function's derivative (up to a constant) is the signal for this substitution.

Answer: Step 2

The choices in Step 1 are the standard ones and are correct. Step 2 misquotes the formula: integration by parts reads $\int udv=uv-\int vdu$, with a minus sign. Correctly, $\int \ln xdx=x\ln x-\int 1dx=x\ln x-x+C$. Step 3 simplifies the wrong expression without adding new mistakes.

Answer: 0.5

Setting $u=\ln x$ gives $du=\frac{1}{x}dx$, and the bounds map to $u=0$ and $u=1$. The integral becomes ${\int }_0^{1}udu=\frac{1}{2}$.

Answer: 2

The radicand is the inner function and its derivative $2x$ is there up to the constant 2. I set $u=x^2+9$, $du=2xdx$, so $xdx=\frac{1}{2}du$, and convert the limits: $x=0$ gives $u=9$, $x=4$ gives $u=25$. The integral becomes $\frac{1}{2}{\int }_9^{25}{u}^{-1/2}du=\frac{1}{2}{\left[2\sqrt{u}\right]}_9^{25}=\sqrt{25}-\sqrt{9}=5-3=2$. Check by differentiating the indefinite version $\sqrt{x^2+9}$: the chain rule gives $\frac{2x}{2\sqrt{x^2+9}}=\frac{x}{\sqrt{x^2+9}}$, the integrand.

Answer: B — $\frac{1}{2}\ln (x^2+4)+C$

The derivative of the denominator is $2x$, and the numerator is half of that, so this is the logarithm form with a constant to fix. With $u=x^2+4$, $du=2xdx$, I have $xdx=\frac{1}{2}du$ and the integral is $\frac{1}{2}\int \frac{du}{u}=\frac{1}{2}\ln |u|+C=\frac{1}{2}\ln (x^2+4)+C$. No absolute value is needed because $x^2+4$ is always positive. Check: differentiating gives $\frac{1}{2}\cdot \frac{2x}{x^2+4}=\frac{x}{x^2+4}$.

Answer: 3

I integrate with $a$ riding along. With $u=x^2$, $du=2xdx$, the limits become $0$ and $a^2$, and the integral is $\frac{1}{2}{\int }_0^{a^2}{e}^{u}du=\frac{1}{2}\left(e^{a^2}-1\right)$. Setting this equal to $\frac{e^9-1}{2}$ gives $e^{a^2}=e^9$, so $a^2=9$ and $a=\pm 3$. The condition $a>0$ leaves $a=3$. Check: $\frac{1}{2}(e^9-1)$ is what the question asked for. The exponential function is one-to-one, which is what allowed me to equate the exponents.

Answer: 9

The amount is the integral of the rate: ${\int }_0^3\frac{20t}{(t^2+1{)}^2}dt$. The denominator holds the inner function and the numerator carries its derivative up to a constant: $u=t^2+1$, $du=2tdt$, so $20tdt=10du$. Limits: $t=0$ gives $u=1$, $t=3$ gives $u=10$. The integral is $10{\int }_1^{10}{u}^{-2}du=10{\left[-\frac{1}{u}\right]}_1^{10}=10\left(-\frac{1}{10}+1\right)=10\cdot \frac{9}{10}=9$. So 9 litres. Plausibility: the rate peaks near $t=0.6$ at under 7 litres per minute and falls to $0.6$ by $t=3$, so a total of 9 over three minutes is reasonable.