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Let $g:[0,2]\rightarrow\mathbb{R}$ be defined by, $$g(x)=\int_0^x(x-t)e^{t}dt.$$

What will be the area between the curve $y=g''(x)$ and the $x$-axis over the interval $[0,2]?$


If we integrate the given integral by leibniz rule,then,

$$g'(x)=(x-x)e^x\frac{d}{dx}(x)-0,$$ which is equal to $0$.

I am not sure whether I am making any mistake while differentiating or not. How can we find the required area?

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Check $g'(x)$ again using the Leibniz integral rule $$ {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left(\int _{a(x)}^{b(x)}f(t,x)\,\mathrm {d} t\right) =\int _{a(x)}^{b(x)}{\frac {\partial f}{\partial x}}\,\mathrm {d} t\,+\,f{\big (}b(x),x{\big )}\cdot b'(x)\,-\,f{\big (}a(x),x{\big )}\cdot a'(x)}, $$

which shows that your calculation of $g'(x)$ is wrong.


How can we find the required area?

Well, you want $$ \int_0^2g''(x)\ dx. $$ Use the fundamental theorem of calculus and the formula for $g'(x)$.

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  • $\begingroup$ I just change the notations to match your definition of $g(x)$. Can you do the calculation now? $\endgroup$ – Jack Jun 1 '16 at 14:30
  • $\begingroup$ But,when $f(x,t)=f(x)$ then, $$\frac{d}{dx}\int_{a(t)}^{b(t)}f(x)dx=f(b(x))⋅b′(x)−f(a(x))⋅a′(x)$$,applying this I am getting the value of the integral as $0$.Can you please explain where actually I am making the mistake? Thank you. $\endgroup$ – P.B. Jun 1 '16 at 14:34
  • $\begingroup$ You are using the wrong formula. In the definition of your $g(x)$, the integrand has two variables, $t$ and $x$. $\endgroup$ – Jack Jun 1 '16 at 14:36
  • $\begingroup$ Now,I got it.Using the formula we will get,$$g'(x)=\int_0^xe^{t}dt$$,and the rest of it will be $0.$ $\endgroup$ – P.B. Jun 1 '16 at 14:52

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