Question

Suppose that \(\int _ { 1 } ^ { 2 } f ( x ) d x = 5 .\) Find each integral. $$(a)\int _ { 1 } ^ { 2 } f ( u ) d u \quad \quad ( \mathbf { b } ) \int _ { 1 } ^ { 2 } \sqrt { 3 } f ( z ) d z$$ $$(c)\int _ { 2 } ^ { 1 } f ( t ) d t \quad$$ $$(d)\int _ { 1 } ^ { 2 } [ - f ( x ) ] d x$$

Short answer

The integral (a) equals to 5. The integral (b) equals to \(5\sqrt{3}\). The integral (c) equals to -5. The integral (d) equals to -5.

Step-by-step solution

Process Integral (a)

For integral (a) \(\int_{1}^{2} f(u) \, du\), we're using a different variable \(u\) instead of \(x\), but the limits and the function are the same. Therefore, by the property of definite integrals stating that changing the variable of integration does not affect the value of it, it equals to \(\int_{1}^{2} f(x) \, dx\) which value is given as 5.

Process Integral (b)

For integral (b) \(\int_{1}^{2} \sqrt{3} \cdot f(z) \, dz\), again a different variable is used which does not change the value of the integral. Moreover, here we have multiplied the function \(f(z)\) by \(\sqrt{3}\). Because constants can be factored out of an integral, the solution will be \(\sqrt{3} \cdot \int_{1}^{2} f(x) \, dx\) or \(\sqrt{3} \cdot 5\).

Process Integral (c)

For integral (c) \(\int_{2}^{1} f(t) \, dt\), the limits of integration are switched compared to the given integral. The definite integral property states that reversing the limits of integration changes the sign of the integral. Therefore, the solution to this integral is \(- \int_{1}^{2} f(x) \, dx\) or \( -5\).

Process Integral (d)

For integral (d) \(\int_{1}^{2} [-f(x)] \, dx\), the function is multiplied by -1. Because constant factors can be factored away from an integral, the solution to this integral is \(-1 \cdot \int_{1}^{2} f(x) \, dx\) or \(-5\).