Suppose f^'' is continuous on [1,3] and f(1)=2, f(3)=-1, f^'(1)=2, and f^'(3)=-1 . Evaluate ∫1^3

Name: Problem 58, Chapter 7: Applied Calculus for the MLSS A Brief Approach
Uploaded: 2021-05-17T09:51:26-08:00
Duration: 5 min 27 s
Channel: Naresh Bagrecha
Description: Suppose f^'' is continuous on [1,3] and f(1)=2, f(3)=-1, f^'(1)=2, and f^'(3)=-1 . Evaluate ∫1^3 x f^''(x) d x.

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Question

Suppose $f^{\prime \prime}$ is continuous on [1,3] and $f(1)=2, f(3)=-1, f^{\prime}(1)=2,$ and $f^{\prime}(3)=-1 .$ Evaluate $\int_{1}^{3} x f^{\prime \prime}(x) d x.$

   Suppose $f^{\prime \prime}$ is continuous on [1,3] and $f(1)=2, f(3)=-1, f^{\prime}(1)=2,$ and $f^{\prime}(3)=-1 .$ Evaluate $\int_{1}^{3} x f^{\prime \prime}(x) d x.$

Applied Calculus for the MLSS A Brief Approach

Soo T. Tan 10th Edition

Chapter 7, Problem 58

Instant Answer

Solved by Expert Naresh Bagrecha

05/17/2021

Step 1

$ Integration by parts states that $\int u d v=u v-\int v d u,$ where $u$ and $v$ are differentiable functions. In this case, we can choose $u=x$ and $d v=f^{\prime \prime}(x) d x.$ Then, we have $d u=d x$ and $v=\int f^{\prime \prime}(x) d x=f^{\prime}(x)+C,$ Show more…

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Suppose $f^{\prime \prime}$ is continuous on [1,3] and $f(1)=2, f(3)=-1, f^{\prime}(1)=2,$ and $f^{\prime}(3)=-1 .$ Evaluate $\int_{1}^{3} x f^{\prime \prime}(x) d x.$

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Key Concepts

Definite Integration

Definite integration involves evaluating the integral of a function over a specific interval. In practice, this often requires calculating an antiderivative and then taking the difference of its values at the upper and lower limits, thereby measuring the net area under the curve.

Integration by Parts

Integration by parts is a fundamental technique for integrating the product of two functions. It stems from the product rule for differentiation and allows one to replace a complicated integral with potentially simpler pieces by choosing parts of the integrand to differentiate and integrate respectively.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration. It states that if a function is continuous on a closed interval, then its definite integral over that interval can be computed using its antiderivative evaluated at the boundary points.

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