
If \[\int\limits_{ - 1}^4 {f\left( x \right)} dx = 4\] and \[\int\limits_2^4 {\left( {3 - f\left( x \right)} \right)} dx = 7\] , then what is the value of \[\int\limits_2^{ - 1} {f\left( x \right)} dx\]?
A. 2
B. \[ - 3\]
C. \[ - 5\]
D. None of these
Answer
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Hint: Here, two equations of the definite integrals are given. First, solve the given integral \[\int\limits_2^4 {\left( {3 - f\left( x \right)} \right)} dx = 7\] and calculate the value of \[\int\limits_2^4 {f\left( x \right)} dx\]. Then, simplify the integral by applying the formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\]. Similarly, simplify \[\int\limits_{ - 1}^4 {f\left( x \right)} dx = 4\] by using the formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\]. After that, simplify the integral \[\int\limits_2^{ - 1} {f\left( x \right)} dx\] by using the integration rule \[\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx\]. In the end, solve the equations and get the required answer.
Formula Used:\[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\]
\[\int\limits_a^b n dx = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\], where \[n\] is a number.
Complete step by step solution:The given equations of the definite integrals are \[\int\limits_{ - 1}^4 {f\left( x \right)} dx = 4\] and \[\int\limits_2^4 {\left( {3 - f\left( x \right)} \right)} dx = 7\].
Let’s solve the integral \[\int\limits_2^4 {\left( {3 - f\left( x \right)} \right)} dx = 7\].
Apply the integration rule \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\] on the left-hand side.
\[\int\limits_2^4 3 dx - \int\limits_2^4 {f\left( x \right)} dx = 7\]
Solve the first term by applying the integration formula \[\int\limits_a^b n dx = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\].
\[\int\limits_2^4 3 dx - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow \left[ {3x} \right]_2^4 - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow 3\left( {4 - 2} \right) - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow 6 - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow \int\limits_2^4 {f\left( x \right)} dx = - 1\]
Apply the standard formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Thus, \[\int\limits_2^4 {f\left( x \right)} dx = F\left( 4 \right) - F\left( 2 \right) = - 1\]. \[.....\left( 1 \right)\]
Simplify the given equation \[\int\limits_{ - 1}^4 {f\left( x \right)} dx = 4\].
Apply the standard formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
We get,
\[\int\limits_{ - 1}^4 {f\left( x \right)} dx = F\left( 4 \right) - F\left( { - 1} \right) = 4\] \[.....\left( 2 \right)\]
We have to calculate the value of the integral \[\int\limits_2^{ - 1} {f\left( x \right)} dx\].
Apply the integration rule\[\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx\].
Then, \[\int\limits_2^{ - 1} {f\left( x \right)} dx = - \int\limits_{ - 1}^2 {f\left( x \right)} dx\]
Apply the standard formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\] on the right-hand side.
\[\int\limits_2^{ - 1} {f\left( x \right)} dx = - \left[ {F\left( 2 \right) - F\left( { - 1} \right)} \right]\]
\[ \Rightarrow \int\limits_2^{ - 1} {f\left( x \right)} dx = F\left( { - 1} \right) - F\left( 2 \right)\]
Now to calculate the value of \[\int\limits_2^{ - 1} {f\left( x \right)} dx\] subtract the equation \[\left( 2 \right)\] from the equation \[\left( 1 \right)\].
\[\int\limits_{ - 1}^2 {f\left( x \right)} dx = \left[ {F\left( 4 \right) - F\left( 2 \right)} \right] - \left[ {F\left( 4 \right) - F\left( { - 1} \right)} \right]\]
\[ \Rightarrow \int\limits_{ - 1}^2 {f\left( x \right)} dx = - 1 - 4\]
\[ \Rightarrow \int\limits_{ - 1}^2 {f\left( x \right)} dx = - 5\]
Option ‘C’ is correct
Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Sometimes they also add integration constant \[c\] in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.
Formula Used:\[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\]
\[\int\limits_a^b n dx = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\], where \[n\] is a number.
Complete step by step solution:The given equations of the definite integrals are \[\int\limits_{ - 1}^4 {f\left( x \right)} dx = 4\] and \[\int\limits_2^4 {\left( {3 - f\left( x \right)} \right)} dx = 7\].
Let’s solve the integral \[\int\limits_2^4 {\left( {3 - f\left( x \right)} \right)} dx = 7\].
Apply the integration rule \[\int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx - \int\limits_a^b {g\left( x \right)} dx\] on the left-hand side.
\[\int\limits_2^4 3 dx - \int\limits_2^4 {f\left( x \right)} dx = 7\]
Solve the first term by applying the integration formula \[\int\limits_a^b n dx = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\].
\[\int\limits_2^4 3 dx - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow \left[ {3x} \right]_2^4 - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow 3\left( {4 - 2} \right) - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow 6 - \int\limits_2^4 {f\left( x \right)} dx = 7\]
\[ \Rightarrow \int\limits_2^4 {f\left( x \right)} dx = - 1\]
Apply the standard formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Thus, \[\int\limits_2^4 {f\left( x \right)} dx = F\left( 4 \right) - F\left( 2 \right) = - 1\]. \[.....\left( 1 \right)\]
Simplify the given equation \[\int\limits_{ - 1}^4 {f\left( x \right)} dx = 4\].
Apply the standard formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
We get,
\[\int\limits_{ - 1}^4 {f\left( x \right)} dx = F\left( 4 \right) - F\left( { - 1} \right) = 4\] \[.....\left( 2 \right)\]
We have to calculate the value of the integral \[\int\limits_2^{ - 1} {f\left( x \right)} dx\].
Apply the integration rule\[\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx\].
Then, \[\int\limits_2^{ - 1} {f\left( x \right)} dx = - \int\limits_{ - 1}^2 {f\left( x \right)} dx\]
Apply the standard formula \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\] on the right-hand side.
\[\int\limits_2^{ - 1} {f\left( x \right)} dx = - \left[ {F\left( 2 \right) - F\left( { - 1} \right)} \right]\]
\[ \Rightarrow \int\limits_2^{ - 1} {f\left( x \right)} dx = F\left( { - 1} \right) - F\left( 2 \right)\]
Now to calculate the value of \[\int\limits_2^{ - 1} {f\left( x \right)} dx\] subtract the equation \[\left( 2 \right)\] from the equation \[\left( 1 \right)\].
\[\int\limits_{ - 1}^2 {f\left( x \right)} dx = \left[ {F\left( 4 \right) - F\left( 2 \right)} \right] - \left[ {F\left( 4 \right) - F\left( { - 1} \right)} \right]\]
\[ \Rightarrow \int\limits_{ - 1}^2 {f\left( x \right)} dx = - 1 - 4\]
\[ \Rightarrow \int\limits_{ - 1}^2 {f\left( x \right)} dx = - 5\]
Option ‘C’ is correct
Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Sometimes they also add integration constant \[c\] in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.
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