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Question: Evaluate \(\int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} \)....

Evaluate ∫−231x+7dx\int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} .

Explanation

Solution

We will use the formula of integration of inverse function to get the integration of the function given inside the integral and then just put in the upper limit and subtract the value after putting the lower limit in it.

Complete step-by-step answer:
We have with us ∫−231x+7dx\int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} .
Let f(x)=1x+7f(x) = \dfrac{1}{{x + 7}}.
We know that ∫dxx+a=ln⁡∣x+a∣+C\int {\dfrac{{dx}}{{x + a}} = \ln |x + a| + C}
Hence, ∫f(x)dx=∫dxx+7=ln⁡∣x+7∣+C\int {f(x)dx} = \int {\dfrac{{dx}}{{x + 7}}} = \ln |x + 7| + C
Now, if we have a definite integral, we do not have any arbitrary constant and we just subtract the value of the function with lower limit from the value of the function with upper limit. In simpler words, if ∫f(x)dx=F(x)+C\int {f(x)} dx = F(x) + C, then we will have: ∫abf(x)dx=F(b)−F(a)\int\limits_a^b {f(x)dx} = F(b) - F(a).
Hence, ∫−23f(x)dx=F(3)−F(−2)\int\limits_{ - 2}^3 {f(x)dx} = F(3) - F( - 2), where f(x)=1x+7f(x) = \dfrac{1}{{x + 7}} and F(x)=ln⁡∣x+7∣F(x) = \ln |x + 7|.
Hence, ∫−231x+7dx=ln⁡∣3+7∣−ln⁡∣−2+7∣\int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} = \ln |3 + 7| - \ln | - 2 + 7|
Simplifying the RHS, we will get:-
⇒∫−231x+7dx=ln⁡∣10∣−ln⁡∣5∣\Rightarrow \int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} = \ln |10| - \ln |5|
Now, we can simplify the values in modulus function as well to obtain:-
⇒∫−231x+7dx=ln⁡10−ln⁡5\Rightarrow \int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} = \ln 10 - \ln 5
Now, we will use the formula: ln⁡a−ln⁡b=ln⁡ab\ln a - \ln b = \ln \dfrac{a}{b}
⇒∫−231x+7dx=ln⁡105\Rightarrow \int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} = \ln \dfrac{{10}}{5}
Simplifying the RHS, we will get:-
⇒∫−231x+7dx=ln⁡2\Rightarrow \int\limits_{ - 2}^3 {\dfrac{1}{{x + 7}}dx} = \ln 2.

Hence, the required value of the given integral is ln(2).

Note: The students must remember the fact that the definite integral never consists of any arbitrary constant whereas indefinite integral must never be written without its constant. Let us look at a few examples to see why we have that constant in the indefinite integration.
Let us say we have f’(x) = 2. Now, let us find its indefinite integration, we will get:-
∫f′(x)dx=∫2dx=2x+C\int {f'(x)dx = } \int {2dx = } 2x + C.
Now, if we take the function to be 2x + 5, it will have f’(x) = 2 or even if we take the function to be 2x + 10, then f’(x) = 2. Therefore, while integrating without upper and lower limits, we need to take an arbitrary constant as well which on taking derivatives has lost its identity completely.
Definite integrals are used in too many concepts, like if we want to find the area under any curve within some bounds, we will use definite integral to calculate it.