
If $\dfrac{d}{d x} f(x)=g(x)$ then the value of $\int_{a}^{b} f(x) g(x) d x$ is
A .$f(b)-f(a)$
B .$g(b)-g(a)$
C .$\quad \dfrac{1}{2}\left[\{g(b)\}^{2}-\{g(a)\}^{2}\right]$
D. $\dfrac{1}{2}\left[\{f(b)\}^{2}-\{f(a)\}^{2}\right]$
Answer
163.5k+ views
Hint: To solve this problem, we will use the concept of integration by parts. By understanding the relationship between the derivative of a function and the integral of that function, we will be able to evaluate the given integral and find the correct answer.
Formula Used:
\[\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}\]
Complete Step by step Solution:
Given that
$\dfrac{d}{d x} f(x)=g(x)$
We are asked to find $\int_{a}^{b} f(x) g(x) d x$
$=\int_{a}^{b} f(x) f^{\prime}(x) d x$
Let $f(x)=t$
$a \rightarrow b$
$f(a) \rightarrow f(b)$
$=\int_{f(a)}^{f(b)}(t) d t$
$=\left[\dfrac{t^{2}}{2}\right]_{f(a)}^{f(b)}$
$=\dfrac{[f(b)]^{2}-[f(a)]^{2}}{2}$
So the correct answer is option(D)
Note:The key to solving this problem is understanding the fundamental theorem of calculus and the concept of integration by parts. By using these concepts and the given information, we can find the correct answer. It is important to remember that the value of $\int_{a}^{b} f(x) g(x) dx$ is equal to $f(b)g(b) - f(a)g(a)$ by using integration by parts.
Formula Used:
\[\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}\]
Complete Step by step Solution:
Given that
$\dfrac{d}{d x} f(x)=g(x)$
We are asked to find $\int_{a}^{b} f(x) g(x) d x$
$=\int_{a}^{b} f(x) f^{\prime}(x) d x$
Let $f(x)=t$
$a \rightarrow b$
$f(a) \rightarrow f(b)$
$=\int_{f(a)}^{f(b)}(t) d t$
$=\left[\dfrac{t^{2}}{2}\right]_{f(a)}^{f(b)}$
$=\dfrac{[f(b)]^{2}-[f(a)]^{2}}{2}$
So the correct answer is option(D)
Note:The key to solving this problem is understanding the fundamental theorem of calculus and the concept of integration by parts. By using these concepts and the given information, we can find the correct answer. It is important to remember that the value of $\int_{a}^{b} f(x) g(x) dx$ is equal to $f(b)g(b) - f(a)g(a)$ by using integration by parts.
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