Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 245: 34

25

Let I= $\int^{2}_{-3}(4x+7)dx$ Since $\int(4x+7)dx=$$\frac{4x^{2}}{2}+7x= 2x^{2}+7x=F(x)$ Therefore, by the second fundamental theorem, we get $I= F(2)-F(-3)= $$(2\times2^{2}+7\times2)-[2\times(-3^{2})+(7\times(-3))]$ =22-(-3)= 25