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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 :

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

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

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

Applying the Fundamental Theorem of Calculus, the result of the integral is: 90.

The Fundamental Theorem of Calculus states that:

[tex]\int_{a}^{b} F^{\prime}(t) dt = F(b) - F(a)[/tex]

For this problem, we have that:

[tex]f(t) = F^{\prime}(t)[/tex]

[tex]F(t) = 7t^2 + 3t[/tex]

Then, applying the Theorem:

[tex]\int_{2}^{4} f(t) dt = \int_{2}^{4} F^{\prime}(t) dt = F(4) - F(2)[/tex]

[tex]F(4) = 7(4^2) + 3(4) = 124[/tex]

[tex]F(2) = 7(2^2) + 3(2) = 34[/tex]

[tex]F(4) - F(2) = 124 - 34 = 90[/tex]

The result of the integral is of 90.

A similar problem is given at https://brainly.com/question/13755352

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