Question

Let f(t)=F′(t), with F(t)=7t2+3t, and let a=2 and b=4. Write the integral ∫ baf(t).dt and evaluate it using the Fundamental Theorem of Calculus.

Answer 1

Answer:

the integral I = 90

Step-by-step explanation:

for  

F(t)=7t²+3t and f(t)=F′(t)

then for a=2 and b=4 and according the Fundamental Theorem of Calculus

$I=\int\limits^b_ a {f(t)} \, dt = \int\limits^b_a {F'(t)} \, dt = F(b) - F(a)$

then

F(b) = F(t=4) =  7*4²+3*4 = 124

F(a) = F(t=2) =  7*2²+3*2 = 34

thus

I = F(b)-F(a)= 124 - 34 = 90

I = 90