Question

Let g(x)=Intragal from 0 to x f(t) dt, where r is the function whos graph is shown.
Picture attached with question and graphs. As well as points needed

Answer 1

If

$\displaystyle g(x) = \int_0^x f(t) \, dt$

then g(x) gives the signed area under f(x) over a given interval starting at 0.

In particular,

$\displaystyle g(0) = \int_0^0 f(t) \, dt = 0$

since the integral of any function over a single point is zero;

$\displaystyle g(4) = \int_0^4 f(t) \, dt = 8$

since the area under f(x) over the interval [0, 4] is a right triangle with length and height 4, hence area 1/2 • 4 • 4 = 8;

$\displaystyle g(8) = \int_0^8 f(t) \, dt = 0$

since the area over [4, 8] is the same as the area over [0, 4], but on the opposite side of the t-axis;

$\displaystyle g(12) = \int_0^{12} f(t) \, dt = -8$

since the area over [8, 12] is the same as over [4, 8], but doesn't get canceled;

$\displaystyle g(16) = \int_0^{16} f(t) \, dt = 0$

since the area over [12, 16] is the same as over [0, 4], and all together these four triangle areas cancel to zero;

$\displaystyle g(20) = \int_0^{20} f(t) \, dt = 24$

since the area over [16, 20] is a trapezoid with "bases" 4 and 8, and "height" 4, hence area (4 + 8)/2 • 4 = 24;

$\displaystyle g(24) = \int_0^{24} f(t) \, dt = 64$

since the area over [20, 24] is yet another trapezoid, but with bases 8 and 12, and height 4, hence area (8 + 12)/2 • 4 = 40, which we add to the previous area.