# 4. a) Evaluate integral_0^infinity e^{-5} sin(7 x) d x b) Suppose that a greater than 0 and b is...

## Question:

4. a) Evaluate {eq}\int\limits_0^{\infty} e^{-5} \sin(7 x) d x {/eq}

b) Suppose that {eq}a > 0 \ and \ b {/eq} is any real number. Determine a formula for {eq}\int\limits_0^{\infty} e^{-a} \sin ( b x) d x {/eq}

5. Explain what is wrong with this statement {eq}\int\limits_0^\pi \sec^2 (x) d x = \tan \pi - \tan 0 = 0 {/eq}

## Integration.

We can also find the integration problem by Trapezoidal's rule or Simpson's rule.

In integration, limits define the area under the curve.

The formula is:

{eq}\displaystyle\int sinx\ dx=-cosx+c\\\\ \displaystyle\int sec^2x\ dx=tanx+c\\\\ {/eq}

## Answer and Explanation:

We have to evaluate the given integrals:

**Part 4A.)**

{eq}\displaystyle\int _0^{\infty}e^{-5}sin(7x)dx\\\\ {/eq}

Taking:

{eq}b\rightarrow \infty\\\\ {/eq}

We may write the equation as:

{eq}\lim_{b\rightarrow \infty}\displaystyle\int _0^be^{-5}sin(7x)dx\\\\ =\lim_{b\rightarrow \infty}e^{-5}\displaystyle\int _0^bsin(7x)dx\\\\ {/eq}

Let:

{eq}7x=z\\\\ 7\ dx=dz\\\\ dx=\dfrac{dz}{7}\\\\ {/eq}

Substituting for the values of limits:

When {eq}x=0\\\\ {/eq}

{eq}z=0\\\\ {/eq}

When {eq}x=b\\\\ {/eq}

{eq}z=7b\\\\ {/eq}

Therefore integrating:

{eq}=\lim_{b\rightarrow \infty}e^{-5}\displaystyle\int _0^{7b}sin(z)\cdot \dfrac{dz}{7}\\\\ =\lim_{b\rightarrow \infty}\dfrac{e^{-5}}{7}\left [ -cos(z) \right ]_0^{2b}\\\\ =\dfrac{-e^{-5}}{7}\left [ cos(7b)-cos(0) \right ]\\\\ =\infty\\\\ {/eq}

The given integration is improper integral and it is diverging.

**Part 4B.)**

{eq}\displaystyle\int _0^{\infty}e^{-a}sin(bx)dx\\\\ {/eq}

Taking:

{eq}b\rightarrow \infty\\\\ {/eq}

We may write the equation as:

{eq}=\lim_{t\rightarrow \infty}\displaystyle\int _0^{t}e^{-a}sin(bx)dx\\\\ {/eq}

Let:

{eq}bx=z\\\\ b\ dx=dz\\\\ dx=\dfrac{dz}{b}\\\\ {/eq}

Substituting for the values of limits:

When {eq}x=0\\\\ {/eq}

{eq}z=0\\\\ {/eq}

When {eq}x=t\\\\ {/eq}

{eq}z=bt\\\\ {/eq}

Hence integrating:

{eq}=lim{b\rightarrow \infty}\displaystyle\int _0^{bt}e^{-a}sin(z)\cdot \dfrac{dz}{b}\\\\ =\dfrac{e^{-a}}{b}\left [ -cos(z) \right ]_0^{bt}\\\\ =\dfrac{e^{-a}}{b}\left [ cos(bt)-0 \right ]\\\\ =\infty\\\\ {/eq}

**Part 5.)**

{eq}\displaystyle\int _0^{\pi}sec^2(x)dx\\\\ {/eq}

As we know:

{eq}sec^{x}=\dfrac{1}{cos^2x}\\\\ {/eq}

Taking the denominator as 0 we get:

{eq}sec^2x=0\\\\ cosx=0\\\\ x=\dfrac{\pi}{2}\\\\ {/eq}

Hence integrating by substituting the values:

{eq}\displaystyle\int _0^{\frac{\pi}{2}}sec^2x\ dx\\\\ =\left [ tan(x) \right ]_0^{\frac{\pi}{2}}\\\\ =\left [ tan\left ( \dfrac{\pi}{2} \right )-tan(0) \right ]\\\\ =\infty\\\\ {/eq}

This is the correct answer.

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13