We read your photo as:

Evaluate the definite integral integral01 t (t)\,dt using integration by parts.

Mathematics

ScanSolve AI Answer

Answer

1−1e1 - \frac{1}{e}

Step-by-step solution

9 steps

To solve the definite integral ∫01tcosh⁡(t) dt\int_0^1 t \cosh(t)\,dt, we will use integration by parts.

  1. Recall the integration by parts formula.

    The formula for integration by parts is ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du.

  2. Choose uu and dvdv.

    Let u=tu = t and dv=cosh⁡(t) dtdv = \cosh(t)\,dt.

  3. Find dudu and vv.

    Differentiate uu to find dudu:

    du=dtdu = dt

    Integrate dvdv to find vv:

    v=∫cosh⁡(t) dt=sinh⁡(t)v = \int \cosh(t)\,dt = \sinh(t)

  4. Apply the integration by parts formula to the indefinite integral.

    ∫tcosh⁡(t) dt=tsinh⁡(t)−∫sinh⁡(t) dt\int t \cosh(t)\,dt = t \sinh(t) - \int \sinh(t)\,dt

  5. Solve the remaining integral.

    ∫sinh⁡(t) dt=cosh⁡(t)\int \sinh(t)\,dt = \cosh(t)

  6. Substitute the result back into the expression from Step 4.

    The indefinite integral is:

    ∫tcosh⁡(t) dt=tsinh⁡(t)−cosh⁡(t)+C\int t \cosh(t)\,dt = t \sinh(t) - \cosh(t) + C

  7. Evaluate the definite integral using the limits from 0 to 1.

    ∫01tcosh⁡(t) dt=[tsinh⁡(t)−cosh⁡(t)]01\int_0^1 t \cosh(t)\,dt = \left[ t \sinh(t) - \cosh(t) \right]_0^1

    Apply the Fundamental Theorem of Calculus:

    =(1⋅sinh⁡(1)−cosh⁡(1))−(0⋅sinh⁡(0)−cosh⁡(0))= (1 \cdot \sinh(1) - \cosh(1)) - (0 \cdot \sinh(0) - \cosh(0))

  8. Calculate the values of sinh⁡(1)\sinh(1), cosh⁡(1)\cosh(1), sinh⁡(0)\sinh(0), and cosh⁡(0)\cosh(0).

    Recall that sinh⁡(x)=ex−e−x2\sinh(x) = \frac{e^x - e^{-x}}{2} and cosh⁡(x)=ex+e−x2\cosh(x) = \frac{e^x + e^{-x}}{2}.

    sinh⁡(1)=e1−e−12\sinh(1) = \frac{e^1 - e^{-1}}{2} cosh⁡(1)=e1+e−12\cosh(1) = \frac{e^1 + e^{-1}}{2} sinh⁡(0)=e0−e−02=1−12=0\sinh(0) = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0 cosh⁡(0)=e0+e−02=1+12=1\cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1

  9. Substitute these values into the definite integral expression.

    =(e−e−12−e+e−12)−(0⋅0−1)= \left( \frac{e - e^{-1}}{2} - \frac{e + e^{-1}}{2} \right) - (0 \cdot 0 - 1) =(e−e−1−e−e−12)−(−1)= \left( \frac{e - e^{-1} - e - e^{-1}}{2} \right) - (-1) =(−2e−12)+1= \left( \frac{-2e^{-1}}{2} \right) + 1 =−e−1+1= -e^{-1} + 1 =1−1e= 1 - \frac{1}{e}

    The final answer is 1−1e1 - \frac{1}{e}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QIntegrate integral1^e x x dx using integration by parts.QUsing integration by parts, evaluate the integral integral01/2 theta (3) dtheta.QEvaluate the definite integral.QRewrite the integral and apply integration by parts.QEvaluate the following integrals: I. ∫(6t⁴+4t³−t²+1)dt II. ∫(3x²−6x−2)dx. Evaluate the following by…
Browse by topic:Calculus
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp