Question

Find the area inside a function f(t) = \( \frac{t}{(t+3)(t+2)} dt\) from t = -1 to 0.

(a) 4 ln⁡(3) – 5ln⁡(2)

(b) 3 ln⁡(3)

(c) 3 ln⁡(3) – 4ln⁡(2)

(d) 3 ln⁡(3) – 5 ln⁡(2)

The question was posed to me in an online quiz.

Question is from Improper Integrals topic in section Integral Calculus of Engineering Mathematics

Answer 1

Correct choice is (d) 3 ln⁡(3) – 5 ln⁡(2)

Explanation: Now, F(t)=\(\int \frac{t}{(t+3)(t+2)} dt\)

F(t)=\(\int \frac{t}{(t+3)(t+2)} dt\)

=\(\int [\frac{3}{t+3}-\frac{2}{t+2}]dx\)

=\(\int [\frac{3}{t+3}]dx-\int [\frac{2}{t+2}]dx\)

=3 ln⁡(t+3)-2ln⁡(t+2)

Now area inside a function is,  F(0) – F(-1),

hence, F(0)-F(-1)=3 ln⁡(3)-2 ln⁡(2)-3 ln⁡(2)+2 ln⁡(1)=3 ln⁡(3)-5ln⁡(2)