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Question: How do you find the definite integral for: \(10\sin xdx\) for the intervals \(\left[ 0,\pi \right]\)...

How do you find the definite integral for: 10sinxdx10\sin xdx for the intervals [0,π]\left[ 0,\pi \right]?

Explanation

Solution

For this problem they have asked to calculate the definite integral of the given function in the given interval. So, we will first calculate the indefinite integral of the given function by using the integration formulas af(x)dx=af(x)dx\int{af\left( x \right)dx}=a\int{f\left( x \right)dx} where aa is the constant and sinxdx=cosx+C\int{\sin xdx=-\cos x+C}. By using these formulas, we will calculate the indefinite integral of the given function. To calculate the definite integral now we will apply the given limits to the integration value and simplify them to get the required result.

Complete step by step solution:
Given that, 10sinxdx10\sin xdx.
Let us assume that f(x)=10sinxdxf\left( x \right)=10\sin xdx.
Applying integration on both sides of the above equation, then we will get
f(x)dx=10sinxdx\Rightarrow \int{f\left( x \right)dx}=\int{10\sin xdx}
In the above function we can observe that the value 1010 is a constant and sinx\sin x is the trigonometric function, so we are using the formula af(x)dx=af(x)dx\int{af\left( x \right)dx}=a\int{f\left( x \right)dx} (where aa is the constant) in the above equation, then we will have
f(x)dx=10sinxdx\Rightarrow \int{f\left( x \right)dx}=10\int{\sin xdx}
In the above equation we have the value of sinxdx\int{\sin xdx}. In integration we already have the value of sinxdx\int{\sin xdx} as cosx+C-\cos x+C. Applying this value in the above equation, then we will get
f(x)dx=10(cosx)+C\Rightarrow \int{f\left( x \right)dx}=10\left( -\cos x \right)+C
Up to now we have calculated indefinite integral only. In the problem we have the integral [0,π]\left[ 0,\pi \right]. Applying the limit values to the above calculated integration, then we will get
0πf(x)dx=10[cosx]0π 0πf(x)dx=10[cosπ(cos0)] \begin{aligned} & \int\limits_{0}^{\pi }{f\left( x \right)dx}=10\left[ -\cos x \right]_{0}^{\pi } \\\ & \Rightarrow \int\limits_{0}^{\pi }{f\left( x \right)dx}=10\left[ -\cos \pi -\left( -\cos 0 \right) \right] \\\ \end{aligned}
Applying the known values cosπ=1\cos \pi =-1, cos0=1\cos 0=1 in the above equation, then we will have
0πf(x)dx=10(1+1) 0πf(x)dx=20 \begin{aligned} & \Rightarrow \int\limits_{0}^{\pi }{f\left( x \right)dx}=10\left( 1+1 \right) \\\ & \Rightarrow \int\limits_{0}^{\pi }{f\left( x \right)dx}=20 \\\ \end{aligned}
Hence the value of the definite integral of the given function 10sinxdx10\sin xdx for the interval [0,π]\left[ 0,\pi \right] is 2020.

Note: We can also directly calculate the definite integral without calculating the indefinite integral for this problem. But in some cases, we may have complexes for which we can’t easily find the definite integral directly, so it will be better to calculate an indefinite integral and then apply limits to the calculated value.