Mister Exam

Other calculators

Solving integrals/ (7-2x)*lg(x)

Integral of (7-2x)*lg(x) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
 1000                   
   /                    
  |                     
  |  (7 - 2*x)*log(x) dx
  |                     
 /                      
 100
∫1001000(7−2x)log⁡(x) dx\int\limits_{100}^{1000} \left(7 - 2 x\right) \log{\left(x \right)}\, dx 
Integral((7 - 2*x)*log(x), (x, 100, 1000))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=log⁡(x)u = \log{\left(x \right)}.

      Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

      ∫(−2ue2u+7ueu) du\int \left(- 2 u e^{2 u} + 7 u e^{u}\right)\, du

      1. Integrate term-by-term:

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−2ue2u) du=−2∫ue2u du\int \left(- 2 u e^{2 u}\right)\, du = - 2 \int u e^{2 u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu2 du\int \frac{e^{u}}{2}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                False\text{False}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          2. The integral of a constant times a function is the constant times the integral of the function:

            ∫e2u2 du=∫e2u du2\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu2 du\int \frac{e^{u}}{2}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                False\text{False}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            So, the result is: e2u4\frac{e^{2 u}}{4}

          So, the result is: −ue2u+e2u2- u e^{2 u} + \frac{e^{2 u}}{2}

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫7ueu du=7∫ueu du\int 7 u e^{u}\, du = 7 \int u e^{u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. The integral of the exponential function is itself.

              ∫eu du=eu\int e^{u}\, du = e^{u}

            Now evaluate the sub-integral.

          2. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: 7ueu−7eu7 u e^{u} - 7 e^{u}

        The result is: −ue2u+7ueu+e2u2−7eu- u e^{2 u} + 7 u e^{u} + \frac{e^{2 u}}{2} - 7 e^{u}

      Now substitute uu back in:

      −x2log⁡(x)+x22+7xlog⁡(x)−7x- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x

    Method #2

    1. Rewrite the integrand:

      (7−2x)log⁡(x)=−2xlog⁡(x)+7log⁡(x)\left(7 - 2 x\right) \log{\left(x \right)} = - 2 x \log{\left(x \right)} + 7 \log{\left(x \right)}

    2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫(−2xlog⁡(x)) dx=−2∫xlog⁡(x) dx\int \left(- 2 x \log{\left(x \right)}\right)\, dx = - 2 \int x \log{\left(x \right)}\, dx

        1. Let u=log⁡(x)u = \log{\left(x \right)}.

          Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

          ∫ue2u du\int u e^{2 u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu2 du\int \frac{e^{u}}{2}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                False\text{False}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          2. The integral of a constant times a function is the constant times the integral of the function:

            ∫e2u2 du=∫e2u du2\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu2 du\int \frac{e^{u}}{2}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                False\text{False}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            So, the result is: e2u4\frac{e^{2 u}}{4}

          Now substitute uu back in:

          x2log⁡(x)2−x24\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}

        So, the result is: −x2log⁡(x)+x22- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫7log⁡(x) dx=7∫log⁡(x) dx\int 7 \log{\left(x \right)}\, dx = 7 \int \log{\left(x \right)}\, dx

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(x)=log⁡(x)u{\left(x \right)} = \log{\left(x \right)} and let dv⁡(x)=1\operatorname{dv}{\left(x \right)} = 1.

          Then du⁡(x)=1x\operatorname{du}{\left(x \right)} = \frac{1}{x}.

          To find v(x)v{\left(x \right)}:

          1. The integral of a constant is the constant times the variable of integration:

            ∫1 dx=x\int 1\, dx = x

          Now evaluate the sub-integral.

        2. The integral of a constant is the constant times the variable of integration:

          ∫1 dx=x\int 1\, dx = x

        So, the result is: 7xlog⁡(x)−7x7 x \log{\left(x \right)} - 7 x

      The result is: −x2log⁡(x)+x22+7xlog⁡(x)−7x- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x

    Method #3

    1. Use integration by parts:

      ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=log⁡(x)u{\left(x \right)} = \log{\left(x \right)} and let dv⁡(x)=7−2x\operatorname{dv}{\left(x \right)} = 7 - 2 x.

      Then du⁡(x)=1x\operatorname{du}{\left(x \right)} = \frac{1}{x}.

      To find v(x)v{\left(x \right)}:

      1. Integrate term-by-term:

        1. The integral of a constant is the constant times the variable of integration:

          ∫7 dx=7x\int 7\, dx = 7 x

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−2x) dx=−2∫x dx\int \left(- 2 x\right)\, dx = - 2 \int x\, dx

          1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

          So, the result is: −x2- x^{2}

        The result is: −x2+7x- x^{2} + 7 x

      Now evaluate the sub-integral.

    2. Rewrite the integrand:

      −x2+7xx=7−x\frac{- x^{2} + 7 x}{x} = 7 - x

    3. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

        ∫7 dx=7x\int 7\, dx = 7 x

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫(−x) dx=−∫x dx\int \left(- x\right)\, dx = - \int x\, dx

        1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

        So, the result is: −x22- \frac{x^{2}}{2}

      The result is: −x22+7x- \frac{x^{2}}{2} + 7 x

    Method #4

    1. Rewrite the integrand:

      (7−2x)log⁡(x)=−2xlog⁡(x)+7log⁡(x)\left(7 - 2 x\right) \log{\left(x \right)} = - 2 x \log{\left(x \right)} + 7 \log{\left(x \right)}

    2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫(−2xlog⁡(x)) dx=−2∫xlog⁡(x) dx\int \left(- 2 x \log{\left(x \right)}\right)\, dx = - 2 \int x \log{\left(x \right)}\, dx

        1. Let u=log⁡(x)u = \log{\left(x \right)}.

          Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

          ∫ue2u du\int u e^{2 u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu2 du\int \frac{e^{u}}{2}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                False\text{False}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          2. The integral of a constant times a function is the constant times the integral of the function:

            ∫e2u2 du=∫e2u du2\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu2 du\int \frac{e^{u}}{2}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                False\text{False}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            So, the result is: e2u4\frac{e^{2 u}}{4}

          Now substitute uu back in:

          x2log⁡(x)2−x24\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}

        So, the result is: −x2log⁡(x)+x22- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫7log⁡(x) dx=7∫log⁡(x) dx\int 7 \log{\left(x \right)}\, dx = 7 \int \log{\left(x \right)}\, dx

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(x)=log⁡(x)u{\left(x \right)} = \log{\left(x \right)} and let dv⁡(x)=1\operatorname{dv}{\left(x \right)} = 1.

          Then du⁡(x)=1x\operatorname{du}{\left(x \right)} = \frac{1}{x}.

          To find v(x)v{\left(x \right)}:

          1. The integral of a constant is the constant times the variable of integration:

            ∫1 dx=x\int 1\, dx = x

          Now evaluate the sub-integral.

        2. The integral of a constant is the constant times the variable of integration:

          ∫1 dx=x\int 1\, dx = x

        So, the result is: 7xlog⁡(x)−7x7 x \log{\left(x \right)} - 7 x

      The result is: −x2log⁡(x)+x22+7xlog⁡(x)−7x- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x

  2. Now simplify:

    x(−2xlog⁡(x)+x+14log⁡(x)−14)2\frac{x \left(- 2 x \log{\left(x \right)} + x + 14 \log{\left(x \right)} - 14\right)}{2}

  3. Add the constant of integration:

    x(−2xlog⁡(x)+x+14log⁡(x)−14)2+constant\frac{x \left(- 2 x \log{\left(x \right)} + x + 14 \log{\left(x \right)} - 14\right)}{2}+ \mathrm{constant}

The answer is:

x(−2xlog⁡(x)+x+14log⁡(x)−14)2+constant\frac{x \left(- 2 x \log{\left(x \right)} + x + 14 \log{\left(x \right)} - 14\right)}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                           2                               
 |                           x           2                    
 | (7 - 2*x)*log(x) dx = C + -- - 7*x - x *log(x) + 7*x*log(x)
 |                           2                                
/
∫(7−2x)log⁡(x) dx=C−x2log⁡(x)+x22+7xlog⁡(x)−7x\int \left(7 - 2 x\right) \log{\left(x \right)}\, dx = C - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x
Simplify
The graph
1002003004005006007008009001000-100000005000000
Plot the graph f(x)
The answer [src] 
488700 - 993000*log(1000) + 9300*log(100)
−993000log⁡(1000)+9300log⁡(100)+488700- 993000 \log{\left(1000 \right)} + 9300 \log{\left(100 \right)} + 488700 
=
= 
488700 - 993000*log(1000) + 9300*log(100)
−993000log⁡(1000)+9300log⁡(100)+488700- 993000 \log{\left(1000 \right)} + 9300 \log{\left(100 \right)} + 488700 
488700 - 993000*log(1000) + 9300*log(100)
Numerical answer [src] 
-6327872.90929957
-6327872.90929957

    Use the examples entering the upper and lower limits of integration.

    © Mister Exam – Calculator