∫0π/4sin4tdt{"language":"en","toolbar":"
Celeb G.
Answered question
2022-09-23
$\int_{0}^{π / 4} \sin (4 t) d t$
$\int_{- 1}^{1} x^{n} d x$
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∫0π/4sin4tdt{&quot;language&quot;:&quot;en&quot;,&quot;toolbar&quot;:&quot;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&quot;}∫-11xndx

karton · Accepted Answer

a. To evaluate the integral ∫0π4sin(4t)dt, we can use a basic integration rule.Let&#039;s rewrite the integral:∫0π4sin(4t)dtNow, we can use the integration rule ∫sin(x)dx=−cos(x):=−14cos(4t)|0π4Evaluating the upper and lower limits:=−14cos(π4)−(−14cos(0))Simplifying further:=−14(22)+14(cos(0))=−28+14Therefore, the value of the integral ∫0π4sin(4t)dt is 14−28.b. To evaluate the integral ∫−11xndx, where n is a constant, we can use the power rule of integration.Let&#039;s consider two cases:Case 1: When n≠−1Using the power rule of integration ∫xndx=xn+1n+1, we have:∫−11xndx=xn+1n+1|−11Evaluating the upper and lower limits:=1n+1n+1−(−1)n+1n+1=1n+1−(−1)n+1n+1Simplifying further:=1+(−1)n+1n+1Case 2: When n=−1In this case, we have:∫−11x−1dxThis integral is improper as it results in division by zero at x=0. Therefore, the integral is undefined.

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