Practice

$$\text{Suppose } \int_{a}^{b} f(x)\,dx = 1. \quad \text{Find } \int_{b}^{a} 2f(x)\,dx.$$

$$\int 0\,dx \;=\; ?$$

$$\int_{a}^{b} 0\,dx \;=\; ?$$

$$\text{Suppose } \int_{0}^{2} f(x)\,dx = 5. \quad \text{Find } \int_{2}^{0} \big(3f(x)+4\big)\,dx.$$

Solutions

1)
Use linearity and reversing limits:

$$\int_{a}^{b} kf(x)\,dx = k\int_{a}^{b} f(x)\,dx, \qquad \int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx.$$

So

$$\int_{b}^{a} 2f(x)\,dx = -\,2\int_{a}^{b} f(x)\,dx = -2.$$

2)
Antiderivative of $0$ is a constant:

$$\int 0\,dx = C.$$

3)

$$\int_{a}^{b} 0\,dx = C\Big|_{a}^{b} = C – C = 0.$$

Geometric view: it’s the area under $y=0$ from $x=a$ to $x=b$, which is $0$.

4)

$$\int_{2}^{0} (3f+4)\,dx = -\int_{0}^{2} (3f+4)\,dx = -\Big(3\!\int_{0}^{2} f(x)\,dx + \int_{0}^{2} 4\,dx\Big) = -\big(3\cdot 5 + 4\cdot 2\big) = -23.$$