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?


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 $g'(x)$.

  • $\begingroup$ I just change the notations to match your definition of $g(x)$. Can you do the calculation now? $\endgroup$ – user9464 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 $x$. $\endgroup$ – user9464 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.